LeetCode 刷题随手记 – 第一部分
前 256 题（非会员），仅算法题，的吐槽
https://leetcode.com/problemset/algorithms/
说明
刷题指南
- 不要背题，前面的基本题型需要熟悉
- 不要刷太多题
- 前面的基本题型需要熟悉
- 记模版
背题很容易忘，而且题目有具体的要求，还总会有新的题目。
模板，比如大数、回溯、图的深度优先搜索、动态规划。
一次刷一定量的题，方便类比和补遗。对照课本上的知识点。
总之，为了解决题目，不是单独扯道理。
关于 LeetCode
是针对白板编程设计的题目
手写，
测试
AC
LeetCode 的题目还都是基础题，白板编程的，自己写测试
AC 仍可能不对
不 AC 不一定错，比如 LeetCode 改题目了
AC
- 对特殊值处理可能不对
- 与题目中对实现方式的要求不一致
Leetcode 没有对异常输入的处理
有些课本上的知识点目前的有没有涵盖到的
而且仅算法题的涵盖也有点窄
不要在题目的范围和已有的题目上限制
- 其他方面的基础知识，未知的题目的分析方法
每道题选择一个标签，为了按方法分类
- 题后的标签用来按解法归类
- 只指定一个类别，
比如 Array+DP 的
按题号排列，从学习的角度，归类更好。
解答
使用 C/C++，按题号顺序。
分类学习。
1. Two Sum
- 哈希表，O(N)
- 或者先排序再查找，如果不是返回下标
要点
- 数组遍历：用 for 循环，可以求和，最大值，最大值的下标，元素是否存在。
- 双指针：同向，可以两头，可以嵌套。可以快慢，先后，收缩。
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int,int> h;
for(int i=0;;i++){
int x = nums[i];
if(h.count(x))
return {h[x],i};
h[target-x] = i;
}
}
};
2. Add Two Numbers
- 链表，遍历和尾插
- 大数，加法
要点
- 大数：用一个整数表示大数中一位（或几位），计算加法时产生进位。
- 链表：和数组都是基础的容器。链表是递归的结构，可以递归遍历，尾递归写成循环。可以直接头部插入或尾部插入（记录最后节点）。为处理空链表方便，可以通过额外的头节点。
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode* list = NULL;
ListNode** p = &list;
int s = 0;
while(l1 || l2 || s){
if(l1){
s+ = l1->val;
l1 = l1->next;
}
if(l2){
s+ = l2->val;
l2 = l2->next;
}
*p = new ListNode(s%10);
p = &(*p)->next;
s /= 10;
}
return list;
}
};
3. Longest Substring Without Repeating Characters
- 数组
- 动态规划
- O(N)
要点
- 双指针：一个向前一步，另一个根据条件收缩。该题在遍历时记录最大值。
class Solution {
public:
int lengthOfLongestSubstring(string s) {
int y = 0;
int b[128] = {0};
int l = 0;
for(int i=0;i<s.size();i++){
char c = s[i];
l = max(l,b[c]);
y = max(y,i-l+1);
b[c] = i+1;
}
return y;
}
};
class Solution {
public:
int lengthOfLongestSubstring(string s) {
int y = 0;
vector<int> b(128, -1);
int a = -1;
for(int i=0;i<s.size();i++){
char c = s[i];
a = max(a,b[c]);
y = max(y,i-a);
b[c] = i;
}
return y;
}
};
4. Median of Two Sorted Arrays
- 双数组的查找，找中位数。
- 挺有趣的一道题！
- 算总长度，分奇偶
- 转化为求第 k 个值。
- 从 0 开始是，从 1 开始。
- 有个相关的题
- Find the k-th Smallest Element in the Union of Two Sorted Arrays
- http://articles.leetcode.com/find-k-th-smallest-element-in-union-of/
要点
- 双有序数组：搜索时，可以归并为一个数组执行查找，也可以直接在双数组上查找。
class Solution {
double find(vector<int>& A, vector<int>& B, int k){
int a = 0, b = 0;
for(;;){
if(k==0){
if(!(b<B.size()) || (a<A.size() && A[a]<B[b]))
return A[a];
else
return B[b];
}
int c = (k-1)/2;
if(!(b+c<B.size()) || (a+c<A.size() && A[a+c]<=B[b+c])){
a += c+1;
k -= c+1;
}else{
b += c+1;
k -= c+1;
}
}
}
public:
double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
int length = nums1.size() + nums2.size();
if(length%2)
return find(nums1,nums2,length/2);
else
return (find(nums1,nums2,length/2-1)+find(nums1,nums2,length/2))/2.0;
}
};
double find(int* nums1, int nums1Size, int* nums2, int nums2Size, int k){
for(;;){
if(nums1Size>nums2Size){
int* nums3 = nums2;
nums2 = nums1;
nums1 = nums3;
nums1Size = nums1Size^nums2Size;
nums2Size = nums1Size^nums2Size;
nums1Size = nums1Size^nums2Size;
}else if(nums1Size==0){
return nums2[k-1];
}else if(k==1){
return nums1[0]<nums2[0] ? nums1[0] : nums2[0];
}else{
int k1 = k/2 < nums1Size ? k/2 : nums1Size;
int k2 = k - k1;
if(nums1[k1-1]==nums2[k2-1]){
return nums1[k1-1];
}else if(nums1[k1-1]<nums2[k2-1]){
nums1 += k1;
nums1Size -= k1;
k -= k1;
}else{
nums2 += k2;
nums2Size -= k2;
k -= k2;
}
}
}
}
double findMedianSortedArrays(int* nums1, int nums1Size, int* nums2, int nums2Size) {
int length = nums1Size + nums2Size;
double med = find(nums1, nums1Size, nums2, nums2Size, length/2+1);
if(length%2==0)
med = (med+find(nums1, nums1Size, nums2, nums2Size, length/2))/2;
return med;
}
5. Longest Palindromic Substring
- 动态规划，后面回文子串的题目还会用到
- 从前往后，计算前缀，或者从后往前，计算后缀都可以
- 这题也有别的更好的方法。
class Solution {
public:
string longestPalindrome(string s) {
int len = 0, start;
int n = s.size();
auto f = [&](int j, int k){
while(j>=0 && k<n && s[j]==s[k]){
j--, k++;
}
if(k-j-1>len){
len = k-j-1;
start = j+1;
}
};
for(int i=0; i<n; i++){
f(i, i);
f(i, i+1);
}
return s.substr(start, len);
}
};
class Solution {
public:
string longestPalindrome(string s) {
int len = 0, start;
int n = s.size();
bool d[n][n];
memset(d, 0, sizeof(bool)*(n*n));
for(int i=0; i<n; i++){
for(int j=0;j<=i; j++){
d[j][i] = s[j]==s[i] && (i-j<3 ||d[j+1][i-1]);
if(d[j][i] && (i-j+1) > len){
len = i - j + 1;
start = j;
}
}
}
return s.substr(start, len);
}
};
6. ZigZag Conversion
- 存在下标周期性映射关系
- 分析 r=1,2,3,4,5 的情况
class Solution {
public:
string convert(string s, int numRows) {
if(numRows==1)
return s;
int d = 2*numRows-2;
string y;
for(int i=0; i<numRows; i++){
for(int j=i; j<s.size(); j+=d){
y.push_back(s[j]);
if(i!=0 && i!=numRows-1){
int k = j+d-i*2;
if(k < s.size()){
y.push_back(s[k]);
}
}
}
}
return y;
}
};
// numRows = 5
// 0 8 16
// 1 7 9 15 17
// 2 6 10 14
// 3 5 11 13
// 4 12
// 0 4
// 1 3 5 7
// 2 6
// 0 2
// 1 3
// 0 1 2 3
7. Reverse Integer
Integer
- 32位补码表示有符号整数
- 符号不变，转换结果可能会溢出
要点
- 十进制：用取模和整数除法，可以得到最低位和余下的高位。
- 整型溢出：运算后超出表示范围，可以在运算前，或者运算后检查
class Solution {
public:
int reverse(int x) {
int a = x>0 ? INT_MAX : INT_MIN;
int b = a/10, c = a%10;
int y = 0;
for(;x>=10 || x<=-10; x /= 10)
y = y*10 + x%10;
if(x>0)
return y<b || (y==b && x<=c) ? y*10+x : 0;
else
return y>b || (y==b && x>=c) ? y*10+x : 0;
}
};
class Solution {
public:
int reverse(int x) {
int a = x>=0 ? INT_MAX : INT_MIN;
int b = a/10, c = a%10;
int y = 0;
if(x>=0){
for(;x>=10; x /= 10)
y = y*10 + x%10;
return y<b || (y==b && x<=c) ? y*10+x : 0;
}else{
for(;x<=-10; x /= 10)
y = y*10 + x%10;
return y>b || (y==b && x>=c) ? y*10+x : 0;
}
}
};
这道题 LeetCode 的 Blog 上的解法挺不错的
8. String to Integer (atoi)
- 字符串转整数，实现 C 语言的字符串类库函数
- 可以用
7. Reverse Integer用到的溢出的判断方法- 区分处理正负补码的表示范围
int myAtoi(char* str) {
while(isspace(*str))
str++;
int neg = 0;
switch(*str){
case '-':
neg = 1;
case '+':
str++;
break;
}
if(!isdigit(*str))
return 0;
int a, b;
if(!neg)
a = INT_MAX/10, b = INT_MAX%10;
else
a = -(INT_MIN/10), b = -(INT_MIN%10);
int value = 0;
while(isdigit(*str)){
int digit = *str-'0';
if(value>a || (value==a && digit>b))
return neg ? INT_MIN : INT_MAX;
value = value*10 + digit;
str++;
}
return neg ? -value : value;
}
9. Palindrome Number
- 逐对比较
- 十进制自然数运算
取最高位和最低位
class Solution {
public:
bool isPalindrome(int x) {
if(x<0)
return false;
int c = 1;
while(x/c>=10)
c *= 10;
while(x){
if((x/c)!=(x%10))
return false;
x = (x%c)/10;
c /= 100;
}
return true;
}
};
10. Regular Expression Matching
- 递归/回溯处理字符串后缀
- 注意各种情况的组合
class Solution {
bool _isMatch(const char* s, const char* p) {
if(*p==0){
return *s==0;
}else if(*(p+1)=='*'){
if(*s==0)
return _isMatch(s, p+2);
else if(*s==*p || *p=='.')
return _isMatch(s+1, p) || _isMatch(s, p+2);
else
return _isMatch(s, p+2);
}else{
if(*s==0)
return false;
if(*s==*p || *p=='.')
return _isMatch(s+1, p+1);
else
return false;
}
}
public:
bool isMatch(string s, string p) {
return _isMatch(s.c_str(), p.c_str());
}
};
11. Container With Most Water
- 数组，动态规划，双指针
- 能装最多的水
- 双指针，从两头，次高往最高收缩
class Solution {
public:
int maxArea(vector<int>& height) {
int i = 0, j = height.size()-1;
int m = 0;
while(i<j){
m = max(m, min(height[i],height[j])*(j-i));
if(height[i] < height[j])
i++;
else
j--;
}
return m;
}
};
12. Integer to Roman
- 罗马数字转换，输入为 1 到 3999
class Solution {
public:
string intToRoman(int num) {
int integers[13] = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
char* romans[13] = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
string sb;
for(int i=0; i<13; i++){
for(int j=num/integers[i]; j--;)
sb += romans[i];
num %= integers[i];
}
return sb;
}
};
13. Roman to Integer
- 罗马数字转换
int romanToInt(char* s) {
int integers[13] = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
char* romans[13] = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
int result = 0;
for(int i=0; i<13; i++){
while(!memcmp(s, romans[i], strlen(romans[i]))){
result += integers[i];
s += strlen(romans[i]);
}
}
return result;
}
14. Longest Common Prefix
- 字符串比较，公共前缀
直接按题意实现即可
char* longestCommonPrefix(char** strs, int strsSize) {
char *result;
if(strsSize==0){
result = (char*)malloc(sizeof(char));
result[0] = 0;
return result;
}
int count = 0;
for(;;){
char ch = strs[0][count];
if(ch==0)
break;
for(int j=1; j<strsSize; j++)
if(strs[j][count]!=ch)
goto end_for;
count++;
}
end_for:
result = (char*)malloc(sizeof(char)*(count+1));
memcpy(result,strs[0],sizeof(char)*count);
result[count] = 0;
return result;
}
15. 3Sum
- 排序然后查找
- 这题有重复元素，比如
[-1, -1, 2]，如果最后再去重可能会超时
- 这题有重复元素，比如
- 注意
.size()返回的是 unsigned，若判断<.size()-1会一直循环
或者用哈希表
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
vector<vector<int>> result;
sort(nums.begin(), nums.end());
int n = nums.size();
for(int i=0; i<n-2; i++){
if(i!=0 && nums[i]==nums[i-1])
continue;
int j = i+1, k = nums.size()-1;
while(j < k){
int s = nums[i] + nums[j] + nums[k];
if(s==0){
result.push_back({nums[i], nums[j], nums[k]});
j++, k--;
}else if(s<0){
j++;
}else{
k--;
}
while(j!=i+1 && nums[j]==nums[j-1])
j++;
while(k!=nums.size()-1 && nums[k]==nums[k+1])
k--;
}
}
return result;
}
};
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
vector<vector<int>> result;
sort(nums.begin(), nums.end());
int n = nums.size();
for(int i=0; i<n-2 && nums[i]<=0; i++){
if(i!=0 && nums[i]==nums[i-1])
continue;
int j = i+1, k = nums.size()-1;
while(j < k){
int s = nums[i] + nums[j] + nums[k];
if(s==0){
result.push_back({nums[i], nums[j], nums[k]});
j++, k--;
}else if(s<0){
j++;
}else{
k--;
}
while(j!=i+1 && nums[j]==nums[j-1])
j++;
while(k!=nums.size()-1 && nums[k]==nums[k+1])
k--;
}
}
return result;
}
};
16. 3Sum Closest
Search
- 先排序
- 或者用哈希
class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
int ans = 0, d = INT_MAX;
sort(nums.begin(), nums.end());
if(nums.size()<3)
return 0;
for(int i=0; i<nums.size()-2; i++){
if(i!=0 && nums[i]==nums[i-1])
continue;
int j = i+1, k = nums.size()-1;
while(j < k){
int s = nums[i] + nums[j] + nums[k] - target;
if(abs(s)<d){
d = abs(s);
ans = s + target;
}
if(s==0){
j++, k--;
}else if(s<0){
j++;
}else{
k--;
}
while(j!=i+1 && nums[j]==nums[j-1])
j++;
while(k!=nums.size()-1 && nums[k]==nums[k+1])
k--;
}
}
return ans;
}
};
17. Letter Combinations of a Phone Number
- 递归，回溯，排列组合
要点
回溯：
class Solution {
public:
vector<string> letterCombinations(string digits) {
vector<string> result;
if(digits.size()==0)
return result;
string y = digits;
int n = digits.size();
char b[digits.size()+1];
b[n] = 0;
char* s[10] = {
"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"
};
function<void(int)> f = [&](int i){
if(i==digits.size()){
result.push_back(b);
}else{
for(char *p = s[digits[i]-'0']; *p; p++){
b[i] = *p;
f(i+1);
}
}
};
f(0);
return result;
}
};
18. 4Sum
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int>> ans;
sort(nums.begin(), nums.end());
for(int x:nums)printf("%d ",x);puts("");
int n = nums.size();
for(int i=0; i<n-3; i++){
if(i!=0 && nums[i]==nums[i-1])
continue;
for(int j=i+1; j<n-2; j++){
if(j!=i+1 && nums[j]==nums[j-1])
continue;
int k = j + 1, l = n-1;
while(k<l){
int sum = nums[i]+nums[j]+nums[k]+nums[l];
if(sum==target){
ans.push_back({nums[i],nums[j],nums[k],nums[l]});
k++, l--;
}else if(sum<target){
k++;
}else{
l--;
}
while(k!=j+1 && k<l && nums[k]==nums[k-1])
k++;
while(l!=n-1 && k<l && nums[l]==nums[l+1])
l--;
}
}
}
return ans;
}
};
19. Remove Nth Node From End of List
- 链表，双指针
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* removeNthFromEnd(struct ListNode* head, int n) {
struct ListNode* p = head;
while(n--)
p = p->next;
if(p==NULL)
return head->next;
p = p->next;
struct ListNode* q = head;
while(p){
q = q->next;
p = p->next;
}
q->next = q->next->next;
return head;
}
20. Valid Parentheses
- 栈，先进后出
要点
stack
bool isValid(char* s) {
const char *ps = "([{)]}";
int stack_size = 3;
char *stack = (char*)malloc(sizeof(char)*stack_size);
char *push = stack;
for(char x; x=*s++;){
char* found = strchr(ps, x);
if(found){
int a = found - ps;
if(a<3){
if(push==stack+stack_size){
stack = (char*)realloc(stack,sizeof(char)*(stack_size<<1));
push = stack + stack_size;
stack_size <<= 1;
}
*push++ = ps[a+3];
}else{
if(push<=stack || push[-1]!=x)
return false;
push--;
}
}
}
return push==stack;
}
21. Merge Two Sorted Lists
- 链表的节点操作
- 有序序列的归并
要点：节点题通常会要求不改变元素，但也不一定。
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
ListNode* head = NULL;
ListNode** append = &head;
for(;;){
if(!l1){
*append = l2;
break;
}
if(!l2){
*append = l1;
break;
}
if(l1->val<l2->val){
*append = l1;
l1 = l1->next;
}else{
*append = l2;
l2 = l2->next;
}
append = &(*append)->next;
}
return head;
}
};
22. Generate Parentheses
- 递归，回溯
class Solution {
public:
vector<string> generateParenthesis(int n) {
vector<string> result;
string item(n*2, ' ');
function<void(int,int)> f = [&](int i,int s){
if(i==n*2){
result.push_back(item);
return;
}
if(s > 0){
item[i] = ')';
f(i+1, s-1);
}
if(s < 2*n - i){
item[i] = '(';
f(i+1, s+1);
}
};
f(0, 0);
return result;
}
};
23. Merge k Sorted Lists
- 归并排序，链表
- 两两归并，次数
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) {
struct ListNode* head = NULL;
struct ListNode** append = &head;
for(;;){
if(!l1){
*append = l2;
break;
}
if(!l2){
*append = l1;
break;
}
if(l1->val<l2->val){
*append = l1;
l1 = l1->next;
}else{
*append = l2;
l2 = l2->next;
}
append = &(*append)->next;
}
return head;
}
struct ListNode* mergeKLists(struct ListNode** lists, int listsSize) {
if(listsSize==0)
return NULL;
if(listsSize==1)
return lists[0];
if(listsSize==2)
return mergeTwoLists(lists[0], lists[1]);
return mergeTwoLists(mergeKLists(lists,listsSize/2),mergeKLists(lists+listsSize/2,listsSize-listsSize/2));
}
24. Swap Nodes in Pairs
- 链表
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* swapPairs(ListNode* head) {
auto list = new ListNode(-1);
list->next = head;
auto p = list;
while(p->next){
auto a = p->next;
auto b = a->next;
if(!b)
break;
auto c = b->next;
a->next = c;
b->next = a;
p->next = b;
p = a;
}
return list->next;
}
};
25. Reverse Nodes in k-Group
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseKGroup(ListNode* head, int k) {
auto list = new ListNode(-1);
list->next = head;
auto last = list;
for(;;){
auto list2 = last;
int n = k;
while(n && last->next){
n--;
last = last->next;
}
if(n)
break;
ListNode* prev = last->next;
ListNode* node = list2->next;
for(int i=0; i<k; i++){
auto next = node->next;
node->next = prev;
prev = node;
node = next;
}
last = list2->next;
list2->next = prev;
}
return list->next;
}
};
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* reverseKGroup(struct ListNode* head, int k) {
struct ListNode **append = &head;
for(;;){
struct ListNode **list = append;
int n = k;
while(n && *append){
append = &(*append)->next;
n--;
}
if(n)
break;
struct ListNode *prev = *append;
struct ListNode *node = *list;
for(int i=0; i<k; i++){
struct ListNode *next = node->next;
node->next = prev;
prev = node;
node = next;
}
auto append2 = &(*list)->next;
*list = prev;
append = append2;
}
return head;
}
26. Remove Duplicates from Sorted Array
- 去重，通过排序。
int removeDuplicates(int* nums, int numsSize) {
if(numsSize==0)
return 0;
int *s = nums;
for(int i=1; i<numsSize;i++){
if(nums[i]!=*s){
*++s = nums[i];
}
}
return s - nums + 1;
}
27. Remove Element
- 数组
- 双指针
int removeElement(int* nums, int numsSize, int val) {
int *s = nums;
for(int i=0; i<numsSize; i++){
if(nums[i]!=val)
*s++ = nums[i];
}
return s - nums;
}
28. Implement strStr()
Search
- 库函数 strstr 实现为 O(MN) ，平时需要 O(N) 就用正则表达式生成状态机
- KMP 算法
要点
- 前缀：字符串头部的连续字符构成的子串。该子串可能会在字符串内重复出现。
class Solution {
int* built_table(string& m){
int *T = new int[m.size()];
T[0] = -1;
T[1] = 0;
for(int i=2, j=0; i<m.size(); ){
printf("> %d \n",i);
if(m[i-1] == m[j]){
T[i] = j + 1;
i++, j++;
}else if(j>0){
j = T[j];
}else{
T[i] = 0;
i++;
}
}
return T;
}
int search(string &s, string &m){
if(m.size()==0)
return 0;
auto next = built_table(m);
for(int i=0, j=0; i<s.size();){
if(m[j]==s[i]){
if(j+1==m.size()){
return i-j;
}else{
i++, j++;
}
}else{
if(next[j]!=-1){
j = next[j];
}else{
i++;
j = 0;
}
}
}
return -1;
}
public:
int strStr(string haystack, string needle) {
return search(haystack, needle);
}
};
稍微改一下，上面的删掉，也有其他写法
class Solution {
int* built_table(string& m){
int *T = new int[m.size()];
T[0] = -1;
T[1] = 0;
for(int i=2, j=0; i<m.size(); ){
if(m[i-1] == m[j]){
T[i] = j + 1;
i++, j++;
}else if(j!=0){
j = T[j];
}else{
T[i] = 0;
i++;
}
}
return T;
}
int search(string &s, string &m){
if(m.size()==0)
return 0;
auto next = built_table(m);
for(int i=0, j=0; i<s.size();){
if(m[j]==s[i]){
if(j+1==m.size())
return i-j;
else
i++, j++;
}else{
if(j!=0)
j = next[j];
else
i++;
}
}
return -1;
}
public:
int strStr(string haystack, string needle) {
return search(haystack, needle);
}
};
这样也行
int strStr(char* s1, char* s2) {
int m = strlen(s2);
if(m==0)
return 0;
int n = strlen(s1);
int b[m+1];
int i, j;
i = 0, j = -1;
b[i] = j;
while(i<m){
while(j>=0 && s2[i]!=s2[j])
j = b[j];
i++, j++;
if(s2[i]==s2[j])
b[i] = b[j];
else
b[i] = j;
}
i = j = 0;
while(i<n){
while(j>=0 && s1[i]!=s2[j])
j = b[j];
i++, j++;
if(j==m)
return i-j;
}
return -1;
}
int strStr(char* s1, char* s2) {
int m = strlen(s2);
if(m==0)
return 0;
int n = strlen(s1);
int b[m+1];
int i, j;
i = 0, j = -1;
b[i] = j;
while(i<m){
while(j>=0 && s2[i]!=s2[j])
j = b[j];
i++, j++;
b[i] = j;
}
i = j = 0;
while(i<n){
while(j>=0 && s1[i]!=s2[j])
j = b[j];
i++, j++;
if(j==m)
return i-j;
}
return -1;
}
29. Divide Two Integers
- 整数运算，通过除法和取模
int divide(int dividend, int divisor) {
if(divisor==0)
return INT_MAX;
unsigned a = dividend>=0 ? dividend : -dividend;
unsigned b = divisor>=0 ? divisor : -divisor;
unsigned c = 0;
int n = 0;
for(unsigned i=b; i<a; i<<=1)
n++;
for(int i=n; i>=0; i--){
if((b<<i)<=a){
c += (1<<i);
a -= (b<<i);
}
}
if((dividend>0 && divisor<0) || (dividend<0 && divisor>0))
return -c;
return c>INT_MAX ? INT_MAX : c;
}
30. Substring with Concatenation of All Words
Array
class Solution {
public:
vector<int> findSubstring(string s, vector<string>& words) {
vector<int> ans;
unordered_map<string,int> count;
for(auto& x: words)
count[x]++;
int n = words.at(0).size();
int m = n*words.size();
for(int i=0,e = s.size()-m+1; i<e; i++){
unordered_map<string,int> copy(count);
for(int j=0; j<words.size(); j++){
auto x = s.substr(i+j*n, n);
auto it = copy.find(x);
if(it!=copy.end() && it->second)
copy[x]--;
else
goto out;
}
ans.push_back(i);
out:;
}
return ans;
}
};
31. Next Permutation
- 排列组合，求下一个排列
- 函数不保存额外的状态，每次调用独立
- 三个元素则依次为 123,132,213,231,312,321
- 值的大小按升序排列，123 开始 321 结束
- 每一位的值表示序号，唯一且有限个
- 从右往左找到第一个小于右侧元素的 A
- 这是可以增大的最低位（从右往左）
- A 右侧一定是降序序列
- 没找到（如 321）则逆置整个排列
- 找到则找到比 A 大的最低位 B
- B 一定是 A 右侧比 A 大的最小值
- A, B 互换，低于 A 的位逆置（从降序变成升序）
- 这是可以增大的最低位（从右往左）
要点
- 升序：等长序列之间也可构成偏序关系。
class Solution {
public:
void nextPermutation(vector<int>& nums) {
int n = nums.size();
int l = n-1;
while(l>0 && !(nums[l-1]<nums[l]))
l--;
if(l>0){
int r = n-1;
while(!(nums[l-1]<nums[r]))
r--;
swap(nums[l-1], nums[r]);
}
for(int r=n-1;l<r;l++,r--)
swap(nums[l], nums[r]);
}
};
32. Longest Valid Parentheses
- 可以用动态规划来做
class Solution {
public:
int longestValidParentheses(string s) {
int len = 0, start = -1;
stack<int> st;
for(int i=0; i<s.size(); i++){
if(s[i]=='('){
st.push(i);
}else if(!st.empty()){
st.pop();
if(!st.empty()){
len = max(len, i-st.top());
}else{
len = max(len, i-start);
}
}else{
start = i;
}
}
return len;
}
};
33. Search in Rotated Sorted Array
Search
- 二分查找
int search(int* A, int n, int target) {
int a = 0, b = n;
while(a<b){
int c = a+(b-a)/2;
if(A[c]==target){
return c;
}else if(A[a]<=A[c]){
if(A[a]<=target && target<A[c])
b = c;
else
a = c + 1;
}else{
if(A[c]<target && target<=A[b-1])
a = c + 1;
else
b = c;
}
}
return -1;
}
34. Search for a Range
Search
- 二分查找
class Solution {
int lower(vector<int>& nums, int target){
int a = 0, b = nums.size();
while(a<b){
int c = (a+b)/2;
if(nums[c]<target){
a = c + 1;
}else{
b = c;
}
}
return b;
}
int upper(vector<int>& nums, int target){
int a = 0, b = nums.size();
while(a<b){
int c = (a+b)/2;
if(nums[c]<=target){
a = c + 1;
}else{
b = c;
}
}
return a-1;
}
public:
vector<int> searchRange(vector<int>& nums, int target) {
int low = lower(nums,target);
int high = upper(nums,target);
if(low<=high)
return {lower(nums,target), upper(nums,target)};
else
return {-1, -1};
}
};
35. Search Insert Position
- 注意右边界是否包含
- 以及终止条件，和连续相等元素的处理
int searchInsert(int* nums, int numsSize, int target) {
int a = 0, b = numsSize;
while(a<b){
int c = (a+b)/2;
if(nums[c]==target){
return c;
}else if(nums[c]<target){
a = c + 1;
}else{
b = c;
}
}
return a;
}
36. Valid Sudoku
- 判断数独局面是否正确
三个条件
class Solution {
public:
bool isValidSudoku(vector<vector<char>>& board) {
bool flag[9];
memset(flag, 0, sizeof(bool)*9);
for(int i=0; i<9; i++){
for(int j=0; j<9; j++){
if(board[i][j]=='.')
continue;
if(flag[board[i][j]-'1'])
return false;
flag[board[i][j]-'1'] = true;
}
memset(flag, 0, sizeof(bool)*9);
for(int j=0; j<9; j++){
if(board[j][i]=='.')
continue;
if(flag[board[j][i]-'1'])
return false;
flag[board[j][i]-'1'] = true;
}
memset(flag, 0, sizeof(bool)*9);
for(int j=0; j<9; j++){
int x=(i%3)*3+j%3;
int y=(i/3)*3+j/3;
if(board[x][y]=='.')
continue;
if(flag[board[x][y]-'1'])
return false;
flag[board[x][y]-'1'] = true;
}
memset(flag, 0, sizeof(bool)*9);
}
return true;
}
};
37. Sudoku Solver
- 回溯
- 有其他高效的算法
class Solution {
public:
void solveSudoku(vector<vector<char>>& board) {
unsigned t[3][9] = {0};
for(int y=0; y<9; y++)
for(int x=0; x<9; x++){
int z = x/3 + y/3*3;
if(board[y][x]!='.'){
int b = board[y][x]-'1';
t[0][x] |= (1<<b);
t[1][y] |= (1<<b);
t[2][z] |= (1<<b);
}
}
function<bool(int)> f = [&](int k){
if(k==81){
return true;
}
int x = k%9, y = k/9;
if(board[y][x]!='.')
return f(k+1);
int z = x/3 + y/3*3;
for(int i=0; i<9; i++){
if (t[0][x]&(1<<i)||t[1][y]&(1<<i)||t[2][z]&(1<<i))
continue;
t[0][x] |= (1 << i);
t[1][y] |= (1 << i);
t[2][z] |= (1 << i);
board[y][x] = i + '1';
if(f(k + 1))
return true;;
board[y][x] = '.';
t[0][x] &= ~(1 << i);
t[1][y] &= ~(1 << i);
t[2][z] &= ~(1 << i);
}
return false;
};
f(0);
}
};
38. Count and Say
- 字符串
class Solution {
public:
string countAndSay(int n) {
string s("1");
while(--n){
string s2;
for(auto it=s.begin(); it!=s.end();){
char ch = *it;
auto it2 = find_if(it, s.end(), [ch](char x){return x!=ch;});
int count = distance(it, it2);
s2 += '0'+count;
s2 += ch;
it = it2;
}
swap(s, s2);
}
return s;
}
};
39. Combination Sum
- 排列组合类型的问题，递归/回溯法
- 该类某项可出现任意多次
- 组合问题各组合升序排列
class Solution {
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
vector<vector<int>> result;
int n = candidates.size();
int sum = 0;
vector<int> item;
function<void(int)> loop = [&](int i){
if(sum>target)
return;
if(sum==target){
result.push_back(item);
return;
}
for(int j=i;j<n;j++){
int x = candidates[j];
sum += x;
item.push_back(x);
loop(j);
sum -= x;
item.pop_back();
}
};
loop(0);
return result;
}
};
40. Combination Sum II
- 排列组合类的问题
- 相比
39. Combination Sum这题有重复元素- 重复运算可以用哈希表或者排序来处理
class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
vector<vector<int>> result;
sort(begin(candidates), end(candidates));
const int n = candidates.size();
vector<int> item;
function<void(int,int)> loop = [&](int i, int sum){
if(sum==target){
result.push_back(item);
return;
}
int y = -1;
for(int j=i; j<n; j++){
int x = candidates[j];
if(x==y)
continue;
if(sum+x>target)
break;
item.push_back(x);
loop(j+1, sum+x);
item.pop_back();
y = x;
}
};
loop(0, 0);
return result;
}
};
class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
vector<vector<int>> result;
sort(begin(candidates), end(candidates));
const int n = candidates.size();
vector<int> item;
function<void(int,int)> loop = [&](int i, int sum){
if(sum==target){
result.push_back(item);
return;
}
for(int j=i; j<n; j++){
int x = candidates[j];
if(j!=i && x==candidates[j-1])
continue;
if(sum+x>target)
break;
item.push_back(x);
loop(j+1, sum+x);
item.pop_back();
}
};
loop(0, 0);
return result;
}
};
41. First Missing Positive
- 题目限制挺具体的
- 数组排列为
[1,2,3,...]- 满足
A[i]=i+1 - 不满足则
i和A[i]-1交换
- 满足
class Solution {
public:
int firstMissingPositive(vector<int>& nums) {
int n = nums.size();
for(int i=0; i<n; i++)
while(nums[i]>0 && nums[i]<=n && nums[i]!=nums[nums[i]-1])
swap(nums[i],nums[nums[i]-1]);
for(int i=0; i<n; i++)
if(nums[i]!=i+1)
return i+1;
return n+1;
}
};
42. Trapping Rain Water
Array
- 数组，双指针（下标）
class Solution {
public:
int trap(vector<int>& height) {
int i = 0, j = height.size()-1;
int a = 0;
int h = 0;
while(i<j){
if(height[i]<height[j]){
h = max(h, height[i]);
a += h - height[i];
i++;
}else{
h = max(h, height[j]);
a += h - height[j];
j--;
}
}
return a;
}
};
43. Multiply Strings
整数
- 大数
- 可以 4 个十进制位一个 32 位整型
char* multiply(char* num1, char* num2) {
int n1 = strlen(num1);
int n2 = strlen(num2);
int n3 = n1 + n2;
char *num3 = (char*)calloc(n3+1, sizeof(char));
for(int i=0; i<n1; i++){
int a = num1[n1-i-1]-'0';
int c = 0;
for(int j=0; j<n2; j++){
int b = num2[n2-j-1]-'0';
int k = n3 - i - j - 1;
c += a*b + num3[k];
num3[k] = c%10;
c /= 10;
}
num3[n3-n2-i-1] += c;
}
char *p = num3;
for(int i=0; i<n3-1; i++){
if(*p!=0)
break;
p++;
}
for(char *q=p; q<num3+n3; q++)
*q += '0';
return p;
}
44. Wildcard Matching
- 字符串搜索
- 通配符
- 遇到星号时，记录下位置
class Solution {
bool isMatch(const char* s, const char* p){
bool star = false;
const char *ss=s, *pp=p;
for(;;){
if(*s==0){
while(*p=='*')
p++;
return *p==0;
}else if(*p=='?'){
if(*s==0)
return false;
s++, p++;
}else if(*p=='*'){
while(*++p=='*')
;
star = true;
ss=s, pp=p;
}else if(*p==*s){
s++, p++;
}else if(star){
ss++;
s=ss, p=pp;
}else{
return false;
}
}
}
public:
bool isMatch(string s, string p) {
return isMatch(s.c_str(), p.c_str());
}
};
45. Jump Game II
class Solution {
public:
int jump(vector<int>& nums) {
int step = 0;
int dist = 0;
int arri = 0;
for(int i=0; i<nums.size(); i++){
if(i>arri){
arri = dist;
step += 1;
}
dist = max(dist , nums[i]+i);
}
return step;
}
};
46. Permutations
- 排列组合问题，求全排列
- 递归/回溯
class Solution {
public:
vector<vector<int>> permute(vector<int>& nums) {
vector<vector<int>> result;
function<void(int)> loop = [&](int i){
if(i==nums.size()){
result.push_back(nums);
return;
}
for(int j=i; j<nums.size(); j++){
swap(nums[i],nums[j]);
loop(i+1);
swap(nums[i],nums[j]);
}
};
loop(0);
return result;
}
};
47. Permutations II
- 排列组合，全排列，有重复元素
class Solution {
public:
vector<vector<int>> permuteUnique(vector<int>& nums) {
vector<vector<int>> result;
if(nums.empty())
return result;
unordered_map<int,int> h;
for(int x: nums)
h[x]++;
vector<int> item;
function<void(int)> loop = [&](int i){
if(i==nums.size()){
result.push_back(item);
return;
}
for(auto p: h){
if(p.second){
item.push_back(p.first);
h[p.first]--;
loop(i+1);
h[p.first]++;
item.pop_back();
}
}
};
loop(0);
return result;
}
};
48. Rotate Image
- 二维数组中元素交换
- 区分尺寸奇偶
- 也可以做两次镜像。
class Solution {
public:
void rotate(vector<vector<int>>& matrix) {
int n = matrix.size();
int s = n - 1;
int e = n&1 ? n/2+1 : n/2;
for(int i=0; i<e; i++)
for(int j=0; j<n/2; j++){
int t = matrix[i][j];
matrix[i][j] = matrix[s-j][i];
matrix[s-j][i] = matrix[s-i][s-j];
matrix[s-i][s-j] = matrix[j][s-i];
matrix[j][s-i] = t;
}
}
};
49. Group Anagrams
- 实现 group 函数
可以利用哈希表或者排序，需要自行实现转化为key
class Solution {
public:
vector<vector<string>> groupAnagrams(vector<string>& strs) {
unordered_map<string,vector<string>> h;
for(auto& s:strs){
auto k = s;
sort(begin(k),end(k));
h[k].push_back(s);
}
vector<vector<string>> xs;
for(auto it = h.begin(); it!=h.end(); ++it){
vector<string> x;
for(auto& e: it->second)
x.push_back(e);
xs.push_back(x);
}
return xs;
}
};
50. Pow(x, n)
- 递归
- 无符号数的位移
- C 用操作数的类型
- Java 里用操作符区分
- 分奇偶
- 有点小问题。
要点
位移：无符号位移高位补0，有符号位移补最高位。正数对应除以2。
double myPow(double x, int n)
{
int neg = n<0;
unsigned exp = neg ? -(unsigned)n : n;
double y = 1;
while(exp)
{
if(exp&1)
y *= x;
x *= x;
exp >>= 1;
}
return neg ? 1/y : y;
}
class Solution {
public:
double myPow(double x, int n) {
printf("n: %d\n",n);
if(n==0)
return 1;
if(n==-1)
return 1/x;
double f = myPow(x,n>>1);
return n%2 ? f*f*x : f*f;
}
};
注意 C 的整数除法是向下取整的
51. N-Queens
- 八皇后
- 求全部解
- 回溯
class Solution {
vector<vector<string>> ans;
int n;
vector<bool> f1, f2, f3;
vector<int> path;
void f(int i){
if(i==n){
vector<string> item(n);
for(int i=0; i<n ;i++){
string line(n, '.');
line[path[i]] = 'Q';
swap(item[i], line);
}
ans.push_back(item);
return;
}
for(int j=0; j<n; j++){
if(f1[j] || f2[i+j] || f3[n-i+j])
continue;
path[i] = j;
f1[j] = f2[i+j] = f3[n-i+j] = true;
f(i+1);
f1[j] = f2[i+j] = f3[n-i+j] = false;
}
}
public:
vector<vector<string>> solveNQueens(int n) {
this->n = n;
f1.resize(n, false);
f2.resize(2*n+1, false);
f3.resize(2*n+1, false);
path.resize(n, false);
f(0);
return ans;
}
};
52. N-Queens II
- 八皇后
同上一题，返回解的个数
class Solution {
int count;
int n;
vector<bool> f1, f2, f3;
void f(int i){
if(i==n){
count++;
return;
}
for(int j=0; j<n; j++){
if(f1[j] || f2[i+j] || f3[n-i+j])
continue;
f1[j] = f2[i+j] = f3[n-i+j] = true;
f(i+1);
f1[j] = f2[i+j] = f3[n-i+j] = false;
}
}
public:
int totalNQueens(int n) {
count = 0;
this->n = n;
f1.resize(n, false);
f2.resize(2*n+1, false);
f3.resize(2*n+1, false);
f(0);
return count;
}
};
53. Maximum Subarray
- 数组，动态规划
- 挺有趣的一道题！
class Solution {
public:
int maxSubArray(vector<int>& nums) {
int m = INT_MIN;
int d = 0;
for(int x: nums){
d = max(d+x, x);
m = max(m, d);
}
return m;
}
};
54. Spiral Matrix
- 按照题意实现，小心不要越界
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
int* spiralOrder(int** matrix, int matrixRowSize, int matrixColSize) {
int *result = (int*)malloc(sizeof(int)*(matrixRowSize*matrixColSize));
int *s = result;
int left = 0, right = matrixColSize-1;
int top = 0, bottom = matrixRowSize-1;
for(;left<=right && top<=bottom;){
for(int i=left; i<=right; i++)
*s++ = matrix[top][i];
for(int i=top+1; i<=bottom; i++)
*s++ = matrix[i][right];
if(top==bottom || left==right)
break;
for(int i=right-1; i>=left; i--)
*s++ = matrix[bottom][i];
for(int i=bottom-1; i>=top+1; i--)
*s++ = matrix[i][left];
left++, right--, top++, bottom--;
}
return result;
}
55. Jump Game
- 题目要求判断是否能抵达终点
class Solution {
public:
bool canJump(vector<int>& nums) {
int dist = 0;
for(int i=0; i<=dist; i++){
dist = max(dist, nums[i]+i);
if(dist>=nums.size()-1)
return true;
}
return false;
}
};
56. Merge Intervals
/**
* Definition for an interval.
* struct Interval {
* int start;
* int end;
* Interval() : start(0), end(0) {}
* Interval(int s, int e) : start(s), end(e) {}
* };
*/
class Solution {
public:
vector<Interval> merge(vector<Interval>& intervals) {
vector<Interval> result;
sort(begin(intervals),end(intervals),
[](Interval& a,Interval& b){return a.start<b.start;});
for(auto& x: intervals){
if(result.empty())
result.push_back(x);
else if(x.start <= result.back().end)
result.back().end = max(x.end, result.back().end);
else
result.push_back(x);
}
return result;
}
};
57. Insert Interval
- 一次移除一个元素不好。
/**
* Definition for an interval.
* struct Interval {
* int start;
* int end;
* Interval() : start(0), end(0) {}
* Interval(int s, int e) : start(s), end(e) {}
* };
*/
class Solution {
public:
vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
for(auto it = intervals.begin(); it!=intervals.end();){
if(newInterval.end < it->start){
intervals.insert(it, newInterval);
return intervals;
}else if(newInterval.start <= it->end){
newInterval.start = min(newInterval.start, it->start);
newInterval.end = max(newInterval.end, it->end);
it = intervals.erase(it);
}else{
++it;
}
}
intervals.push_back(newInterval);
return intervals;
}
};
/**
* Definition for an interval.
* struct Interval {
* int start;
* int end;
* Interval() : start(0), end(0) {}
* Interval(int s, int e) : start(s), end(e) {}
* };
*/
class Solution {
public:
vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {
vector<Interval> result;
auto it=intervals.begin();
while(it!=intervals.end()){
if(newInterval.end < it->start){
break;
}else if(newInterval.start <= it->end){
newInterval.start = min(newInterval.start, it->start);
newInterval.end = max(newInterval.end, it->end);
}else{
result.push_back(*it);
}
++it;
}
result.push_back(newInterval);
while(it!=intervals.end()){
result.push_back(*it);
++it;
}
return result;
}
};
58. Length of Last Word
class Solution {
public:
int lengthOfLastWord(string s) {
int y = 0;
int count = 0;
for(auto p = s.c_str();*p;p++){
if(count!=0)
y = 0;
if(*p==' '){
if(count!=0){
y = count;
count = 0;
}
}else
count++;
}
return count ? count : y;
}
};
59. Spiral Matrix II
- 拿之前那题改一下，读变成写
- 对照
54. Spiral Matrix
/**
* Return an array of arrays.
* Note: The returned array must be malloced, assume caller calls free().
*/
int** generateMatrix(int n) {
int **matrix = (int**)malloc(sizeof(int*)*n);
for(int i=0; i<n ;i++)
matrix[i] = (int*)malloc(sizeof(int)*n);
int count = 1;
int left = 0, right = n-1;
int top = 0, bottom = n-1;
for(;left<=right && top<=bottom;){
for(int i=left; i<=right; i++)
matrix[top][i] = count++;
for(int i=top+1; i<=bottom; i++)
matrix[i][right] = count++;
if(top==bottom || left==right)
break;
for(int i=right-1; i>=left; i--)
matrix[bottom][i] = count++;
for(int i=bottom-1; i>=top+1; i--)
matrix[i][left] = count++;
left++, right--, top++, bottom--;
}
return matrix;
}
60. Permutation Sequence
- 排列组合
- n 个元素有 n! 个排列，n-1 个元素有 (n-1)! 个排列
- 取模/整数除法运算是对从 0 开始的整数
char* getPermutation(int n, int k) {
char *s = (char*)malloc(sizeof(char)*(n+1));
s[n] = 0;
for(int i=0; i<n; i++)
s[i] = '1' + i;
int a = 1;
for(int i=2; i<n; i++)
a *= i;
k--;
for(int i=0; k; i++){
int j = k/a;
char t = s[i+j];
for(int p=i+j; p>i; p--)
s[p] = s[p-1];
s[i] = t;
k %= a;
a /= n-1-i;
}
return s;
}
61. Rotate List
- 链表
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* rotateRight(ListNode* head, int k) {
if(!head || !k)
return head;
ListNode** p = &head;
int n = 0;
while(*p){
n++;
p = &(*p)->next;
}
*p = head;
for(int c=n-k%n; c--;)
p = &(*p)->next;
head = *p;
*p = NULL;
return head;
}
};
62. Unique Paths
- 排列组合
- 动态规划
int nchoosek(int n, int k){
long long res = 1;
for(int i=0;i<k;i++)
res *= n-i;
for(int i=1;i<=k;i++)
res /= i;
return res;
}
int uniquePaths(int m, int n) {
return nchoosek(m+n-2, (m<n ? m : n)-1);
}
63. Unique Paths II
- 动态规划，挺有趣的一道题
int uniquePathsWithObstacles(int** obstacleGrid, int obstacleGridRowSize, int obstacleGridColSize) {
int *d = (int*)malloc(sizeof(int)*obstacleGridColSize);
memset(d,0,sizeof(int)*obstacleGridColSize);
d[0] = 1 - obstacleGrid[0][0];
for(int y=0; y<obstacleGridRowSize; y++){
d[0] = d[0] ? 1-obstacleGrid[y][0] : 0;
for(int x=1; x<obstacleGridColSize; x++)
d[x] = obstacleGrid[y][x] ? 0 : (d[x-1]+d[x]);
}
return d[obstacleGridColSize-1];
}
64. Minimum Path Sum
- 动态规划
- 递推关系
D[i][j] = grid[i][j] + min(D[i-1][j], D[i][j-1])
int minPathSum(int** grid, int gridRowSize, int gridColSize) {
int D[gridColSize];
D[0] = grid[0][0];
for(int i=1; i<gridColSize; i++)
D[i] = grid[0][i] + D[i-1];
for(int i=1; i<gridRowSize; i++){
D[0] += grid[i][0];
for(int j=1; j<gridColSize; j++)
D[j] = grid[i][j] + (D[j]<=D[j-1] ? D[j] : D[j-1]);
}
return D[gridColSize-1];
}
65. Valid Number
- 状态机，正则表达式
bool isNumber(char* s) {
char* e;
double a = strtof(s, &e);
if(e==s)
return false;
while(isspace(*e))
e++;
return *e==0;
}
66. Plus One
- 挺有趣的题，大数和排列组合的题目中会用到。
class Solution {
public:
vector<int> plusOne(vector<int>& digits) {
for(int i=digits.size()-1; i>=0; i--){
if(digits[i]<9){
digits[i]++;
return digits;
}
digits[i] = 0;
}
digits.insert(digits.begin(), 1);
return digits;
}
};
class Solution {
public:
vector<int> plusOne(vector<int>& digits) {
vector<int> result(digits.size());
int carry = 1;
for(int i=digits.size()-1; i>=0; i--){
int digit = digits[i] + carry;
result[i] = digit%10;
carry = digit/10;
}
if(carry)
result.insert(result.begin(), carry);
return result;
}
};
67. Add Binary
- 大数
class Solution {
public:
string addBinary(string a, string b) {
string c;
int m = a.size(), n = b.size();
int carry = 0;
int i = m-1, j = n-1;
for(int k=0, e=max(m,n); k<e; k++){
carry += i>=0 && a[i--]=='1';
carry += j>=0 && b[j--]=='1';
c += '0'+(carry&1);
carry >>= 1;
}
if(carry)
c += '1';
reverse(c.begin(), c.end());
return c;
}
};
68. Text Justification
- 细节略多
class Solution {
public:
vector<string> fullJustify(vector<string>& words, int maxWidth) {
vector<string> ans;
int start = 0;
int size = 0;
for(int i=0; i<words.size(); i++){
string line;
size += words[i].size();
int c = i - start - 1;
if(size+c>=maxWidth){
size = words[i].size();
int space = maxWidth;
for(int j=start; j<i; j++)
space -= words[j].size();
line += words[start++];
if(c==0){
line.append(space, ' ');
}else{
for(; start<i; c--){
int r = space/c+!!(space%c);
line.append(r, ' ');
space -= r;
line += words[start++];
}
}
ans.push_back(line);
}
}
if(start<words.size()){
string line;
for(int i=start;i<words.size(); i++){
if(!line.empty())
line += ' ';
line += words[i];
}
line.append(maxWidth-line.size(), ' ');
ans.push_back(line);
}
return ans;
}
};
69. Sqrt(x)
- 二分法
- 牛顿法
int mySqrt(int x) {
int a = 1, b = x;
while(a<=b){
int c = (a+b)/2;
if(c==x/c)
return c;
else if(c<x/c)
a = c+1;
else
b = c-1;
}
return b;
}
70. Climbing Stairs
- 斐波那契数列
- 下标为台阶个数，值为路径个数
- 满足 a_n = a{n-1}+a{n-1}, a_1=1, a_2=2
int climbStairs(int n) {
int a = 0, b = 1;
while(n--){
int c = a+b;
a = b, b = c;
}
return b;
}
71. Simplify Path
- 栈
class Solution {
public:
string simplifyPath(string path) {
vector<string> stack;
auto it = path.begin();
while(it!=path.end()){
++it;
auto end = find(it,path.end(),'/');
string s(it, end);
if(s==".."){
if(!stack.empty())
stack.pop_back();
}else if(!s.empty() && s!="."){
stack.push_back(s);
}
it = end;
}
if(stack.empty())
return "/";
string result;
for(auto& s: stack){
result += "/";
result += s;
}
return result;
}
};
72. Edit Distance
DP
- 编辑距离
- 动态规划，按字符
- 相等
d[i][j] = d[i-1][j-1] - 替换
d[i][j] = d[i-1][j-1]+1 - 添加
d[i][j] = d[i][j-1]+1 - 删除
d[i][j] = d[i-1][j]+1
- 相等
- 写成递归的形式或迭代的形式
动规
class Solution {
public:
int minDistance(string word1, string word2) {
vector<int> d(word2.size()+1);
for(int i=0; i<=word2.size(); i++)
d[i] = i;
for(int i=1; i<=word1.size(); i++){
int c = d[0];
d[0] = i;
for(int j=1; j<=word2.size() ;j++){
int t = d[j];
if(word1[i-1]==word2[j-1])
d[j] = c;
else
d[j] = min(c, min(d[j-1],d[j])) + 1;
c = t;
}
}
return d[word2.size()];
}
};
递归
class Solution {
public:
int minDistance(string word1, string word2) {
map<pair<int,int>,int> h;
function<int(int,int)> d = [&](int i, int j)->int{
if(h.count({i,j}))
return h[{i,j}];
if(i==word1.size())
return word2.size()-j;
if(j==word2.size())
return word1.size()-i;
int a = (word1[i]==word2[j] ? 0: 1)+d(i+1, j+1);
int b = 1 + d(i+1, j);
int c = 1 + d(i, j+1);
return h[{i,j}] = min(a, min(b, c));
};
return d(0,0);
}
};
73. Set Matrix Zeroes
- 为了不另开空间，第一行第一列先扫一遍，然后用来做记录。
void setZeroes(int** matrix, int matrixRowSize, int matrixColSize) {
int row = 1, col = 1;
for(int i=0;i<matrixRowSize;i++)
if(matrix[i][0]==0){
col = 0;
break;
}
for(int i=0; i<matrixColSize; i++)
if(matrix[0][i]==0){
row = 0;
break;
}
for(int i=0;i<matrixRowSize;i++)
for(int j=0; j<matrixColSize; j++)
if(matrix[i][j]==0)
matrix[i][0] = matrix[0][j] = 0;
for(int i=1;i<matrixRowSize;i++)
if(matrix[i][0]==0)
for(int j=0;j<matrixColSize;j++)
matrix[i][j] = 0;
for(int i=1;i<matrixColSize;i++)
if(matrix[0][i]==0)
for(int j=0;j<matrixRowSize;j++)
matrix[j][i] = 0;
if(row==0)
for(int i=0; i<matrixColSize; i++)
matrix[0][i] = 0;
if(col==0)
for(int i=0;i<matrixRowSize;i++)
matrix[i][0] = 0;
}
74. Search a 2D Matrix
- 二分查找
- 这题直接按照一维就可以了。
bool searchMatrix(int** matrix, int m, int n, int target) {
int a = 0, b = m*n;
while(a<b){
int c = (a+b)/2;
int mid = matrix[c/n][c%n];
if(mid==target)
return true;
else if(mid<target)
a = c + 1;
else
b = c;
}
return false;
}
75. Sort Colors
- 题目要求不要用排序函数
- 这道题正好只有三个元素要排序
- 数组遍历过程中往两端插入
class Solution {
public:
void sortColors(vector<int>& nums) {
int l = 0, r = nums.size()-1;
for(int i=l; i<=r;){
switch(nums[i]){
case 0:
swap(nums[i], nums[l++]);
case 1:
i++;
break;
case 2:
swap(nums[i], nums[r--]);
break;
}
}
}
};
76. Minimum Window Substring
- 双指针
- 收缩比扩展好实现
class Solution {
public:
string minWindow(string s, string t) {
int start = 0;
int size = INT_MAX;
int h[128] = {0};
for(char ch: t)
h[ch]++;
int d[128] = {0};
int count = 0;
for(int i=0, j=0;i<s.size(); i++){
if(h[s[i]]){
d[s[i]]++;
if(d[s[i]]<=h[s[i]])
count++;
if(count==t.size()){
while(d[s[j]]>h[s[j]] || h[s[j]]==0)
d[s[j++]]--;
if(i-j+1<size){
size = i-j+1;
start = j;
}
}
}
}
return size==INT_MAX ? "" : s.substr(start,size);
}
};
77. Combinations
- 排列组合问题，组合是全排列中升序的排列
- 该题的序列是从 1 开始的
class Solution {
public:
vector<vector<int>> combine(int n, int k) {
vector<vector<int>> result;
vector<int> item(k);
function<void(int,int)> loop = [&](int i, int x){
if(i==k){
result.push_back(item);
return;
}
for(int j=x;j<=n;j++){
item[i] = j;
loop(i+1,j+1);
}
};
loop(0, 1);
return result;
}
};
78. Subsets
- 排列组合问题
- 二进制位表示集合
class Solution {
public:
vector<vector<int>> subsets(vector<int>& nums) {
vector<vector<int>> result;
for(int i=0, n=(1<<nums.size()); i<n; i++){
vector<int> item;
for(int j=0; j<nums.size(); j++)
if((i>>j)&1)
item.push_back(nums[j]);
result.push_back(item);
}
return result;
}
};
79. Word Search
- 图的深度优先搜索
- 本题中要求路径无环，矩阵按行储存
bool f(char** board, int boardRowSize, int boardColSize, char* word,int x,int y,bool** visited) {
if(*word==0)
return true;
if(x<0 || x>= boardColSize || y<0 || y>=boardRowSize)
return false;
if(visited[y][x])
return false;
if(board[y][x]==*word){
visited[y][x] = true;
bool result = (
f(board, boardRowSize, boardColSize, word+1, x-1, y, visited)||
f(board, boardRowSize, boardColSize, word+1, x+1, y, visited)||
f(board, boardRowSize, boardColSize, word+1, x, y-1, visited)||
f(board, boardRowSize, boardColSize, word+1, x, y+1, visited)
);
visited[y][x] = false;
return result;
}
return false;
}
bool exist(char** board, int boardRowSize, int boardColSize, char* word) {
bool **visited = (bool**)malloc(sizeof(bool*)*boardRowSize);
for(int i=0;i<boardRowSize;i++)
visited[i] = (bool*)calloc(1, sizeof(bool*)*boardColSize);
for(int y=0; y<boardRowSize; y++)
for(int x=0; x<boardColSize; x++)
if(f(board, boardRowSize, boardColSize, word, x, y, visited))
return true;
return false;
}
80. Remove Duplicates from Sorted Array II
- 有序数组，重复元素相邻
int removeDuplicates(int* nums, int numsSize) {
int *p = nums;
int item = -1, count = 0;
for(int i=0; i<numsSize; i++){
if(nums[i]==item){
if(count==2)
continue;
count++;
}else{
item = nums[i];
count = 1;
}
*p++=nums[i];
}
return p - nums;
}
81. Search in Rotated Sorted Array II
Search
- 二分搜索，旋转后的有序数组
- 比之前的题目多了：有重复元素
bool search(int* A, int n, int target) {
int a = 0, b = n;
while(a<b){
int c = a+(b-a)/2;
if(A[c]==target){
return true;
}else if(A[a]==A[c]){
a++;
}else if(A[a]<A[c]){
if(A[a]<=target && target<A[c])
b = c;
else
a = c + 1;
}else{
if(A[c]<target && target<=A[b-1])
a = c + 1;
else
b = c;
}
}
return false;
}
82. Remove Duplicates from Sorted List II
- 链表节点操作
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* deleteDuplicates(ListNode* head) {
ListNode** p = &head;
while(*p){
ListNode* q = (*p)->next;
if(q && q->val==(*p)->val){
q = q->next;
while(q && q->val==(*p)->val)
q = q->next;
*p = q;
}else
p = &(*p)->next;
}
return head;
}
};
83. Remove Duplicates from Sorted List
- 前面有题目是数组，这次是链表。
class Solution {
public:
ListNode* deleteDuplicates(ListNode* head) {
for(auto p = head; p; p = p->next){
auto q = p->next;
while(q && q->val==p->val)
q = q->next;
p->next = q;
}
return head;
}
};
84. Largest Rectangle in Histogram
- 最大矩形
- Stack，大于则入栈，小于则合并，最后合并全部。
class Solution {
public:
int largestRectangleArea(vector<int>& heights) {
int ans = 0;
stack<int> s;
heights.push_back(0);
for(int i=0; i<heights.size();i++){
while(!s.empty() && heights[i]<=heights[s.top()]){
int j = s.top();
s.pop();
int width = s.empty() ? i : i-s.top()-1;
ans = max(ans, heights[j]*width);
}
s.push(i);
}
return ans;
}
};
85. Maximal Rectangle
- 用
84. Largest Rectangle in Histogram改改 - 也可以用动态规划
class Solution {
public:
int maximalRectangle(vector<vector<char>>& matrix) {
int row = matrix.size();
if(row==0)
return 0;
int col = matrix[0].size();
vector<int> heights(col+1, 0);
int ans = 0;
for(int y=0;y<row; y++){
for(int x=0; x<col; x++)
heights[x] = matrix[y][x]=='1' ? heights[x] + 1 : 0;
stack<int> s;
for(int i=0; i<heights.size();i++){
while(!s.empty() && heights[i]<=heights[s.top()]){
int j = s.top();
s.pop();
int width = s.empty() ? i : i-s.top()-1;
ans = max(ans, heights[j]*width);
}
s.push(i);
}
}
return ans;
}
};
86. Partition List
- 链表节点操作
- 链表追加，可以用一个空的头节点
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
typedef struct ListNode Node;
void list_append(Node*** tail, Node* node){
**tail = node;
*tail = &node->next;
}
struct ListNode* partition(struct ListNode* head, int x) {
Node* list1 = NULL;
Node* list2 = NULL;
Node** append1 = &list1;
Node** append2 = &list2;
for(; head; head=head->next)
list_append(head->val<x ? &append1 : &append2, head);
list_append(&append2, NULL);
list_append(&append1, list2);
return list1;
}
87. Scramble String
DP
- 动态规划
class Solution {
public:
bool isScramble(string s1, string s2) {
map<tuple<int,int,int>,bool> m;
function<bool(int,int,int)> f = [&](int a, int b, int c){
//printf("%d %d %d\n",a,b,c);
if(c==1)
return s1[a]==s2[b];
auto t = make_tuple(a, b, c);
auto it = m.find(t);
if(it!=m.end())
return it->second;
for(int i=1; i<c; i++){
if(f(a,b,i) && f(a+i,b+i,c-i))
return m[t] = true;
if(f(a,b+c-i,i) && f(a+i,b,c-i))
return m[t] = true;
}
return m[t] = false;
} ;
if(s1.size()!=s2.size())
return true;
return f(0,0,s1.size());
}
};
88. Merge Sorted Array
- 有序数组归并
- 这题是在数组中插入另一个数组
void merge(int* nums1, int m, int* nums2, int n) {
int* s = nums1 + m + n-1;
int i = m-1, j = n-1;
while(i>=0 && j>=0)
*s-- = nums1[i] > nums2[j] ? nums1[i--] : nums2[j--];
while(j>=0)
*s-- = nums2[j--];
}
89. Gray Code
class Solution {
public:
vector<int> grayCode(int n) {
vector<int> ans;
for(int i=0,e=1<<n; i<e; i++)
ans.push_back((i>>1)^i);
return ans;
}
};
90. Subsets II
- 递归 + 回溯
- 或者自行用栈实现
- 先统计重复元素个数
class Solution {
public:
vector<vector<int>> subsetsWithDup(vector<int>& nums) {
vector<vector<int>> result;
unordered_map<int,int> h;
for(int x: nums)
h[x]++;
vector<int> keys;
for(auto p: h)
keys.push_back(p.first);
vector<int> stack;
stack.push_back(-1);
while(!stack.empty()){
int n = stack.size();
stack.back()++;
if(stack.back()>h[keys[n-1]]){
stack.pop_back();
}else if(n==keys.size()){
vector<int> item;
for(int j=0; j<n; j++)
for(int k=0; k<stack[j]; k++)
item.push_back(keys[j]);
result.push_back(item);
}else{
stack.push_back(-1);
}
}
return result;
}
};
91. Decode Ways
- 回溯，动态规划
class Solution {
public:
int numDecodings(string s) {
int a = 1, b = 1, c = 0;
char p = 0;
for(char ch: s){
c = 0;
if(ch!='0')
c += b;
if((p=='2'&&ch<='6')||(p=='1'))
c += a;
p = ch;
a = b;
b = c;
}
return c;
}
};
92. Reverse Linked List II
- 挺有趣的一道题
- 在逆置链表的基础上改一下
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
ListNode* reverse(ListNode* head){
ListNode* prev = NULL;
while(head){
auto next = head->next;
head->next = prev;
prev = head;
head = next;
}
return prev;
}
public:
ListNode* reverseBetween(ListNode* head, int m, int n) {
ListNode** p = &head;
for(int i=1; i<m; i++)
p = &(*p)->next;
ListNode* first = *p;
ListNode* last = *p;
for(int i=m; i<n; i++)
last = last->next;
ListNode* next = last->next;
last->next = NULL;
*p = reverse(*p);
first->next = next;
return head;
}
};
93. Restore IP Addresses
- 回溯
class Solution {
public:
vector<string> restoreIpAddresses(string s) {
vector<string> result;
vector<int> path;
function<void(int,int)> f = [&](int i, int x){
if(i==s.size()){
if(path.size()==4 && x==0){
char b[s.size()+4];
sprintf(b,"%d.%d.%d.%d",path[0],path[1],path[2],path[3]);
result.push_back(b);
}
return;
}
int y = x*10 + (s[i]-'0');
if(y>=256)
return;
if(y!=0)
f(i+1, y);
if(path.size()<4){
path.push_back(y);
f(i+1, 0);
path.pop_back();
}
};
f(0, 0);
return result;
}
};
94. Binary Tree Inorder Traversal
- 重点问题！
- 二叉树中序遍历
- LNR，根节点在左右节点之间访问
- 对排序二叉树树得到升序序列
- 递归，非递归
- 先访问最左边的节点
- 记录出栈顺序
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
vector<int> result;
stack<TreeNode*> s;
TreeNode* p = root;
while(p || !s.empty()){
if(p){
s.push(p);
p = p->left;
}else{
p = s.top();
s.pop();
result.push_back(p->val);
p = p->right;
}
}
return result;
}
};
95. Unique Binary Search Trees II
- 挺有趣的题目
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
vector<TreeNode*> generateTrees(int begin, int end) {
vector<TreeNode*> result;
if(begin==end)
result.push_back(NULL);
else
for(int i=begin; i<end; i++)
for(auto left: generateTrees(begin,i))
for(auto right: generateTrees(i+1, end)){
auto node = new TreeNode(i);
node->left = left;
node->right = right;
result.push_back(node);
}
return result;
}
public:
vector<TreeNode*> generateTrees(int n) {
if(n==0)
return {};
return generateTrees(1, n+1);
}
};
96. Unique Binary Search Trees
- 动态规划
- 以上一题为基础
- 选一个根节点，个数为左子树乘以右子树，然后求和
class Solution {
public:
int numTrees(int n) {
vector<int> f(n+1);
f[0] = f[1] = 1;
for(int i=2; i<=n; i++)
for(int j=1; j<=i; j++)
f[i] += f[j-1]*f[i-j];
return f[n];
}
};
97. Interleaving String
DP
动态规划
class Solution {
public:
bool isInterleave(string s1, string s2, string s3) {
int l1 = s1.size();
int l2 = s2.size();
int l3 = s3.size();
if(l1+l2!=l3)
return false;
vector<bool> d(l1+1);
d[0] = true;
for(int i=0; i<l1; i++)
d[i+1] = d[i] && s1[i]==s3[i];
for(int i=0; i<l2; i++){
d[0] = d[0] && s2[i]==s3[i];
for(int j=0; j<l1; j++){
char c = s3[i+j+1];
d[j+1] = (d[j] && s1[j]==c) || (d[j+1] && s2[i]==c);
}
}
return d[l1];
}
};
98. Validate Binary Search Tree
- 检查是否为二叉搜索树
- 节点满足大小升序关系
- 二叉树中序遍历
- 递归，非递归实现
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
int *prev;
bool f(TreeNode* node){
if(!node)
return true;
if(!f(node->left))
return false;
if(prev && *prev>=node->val)
return false;
prev = &node->val;
if(!f(node->right))
return false;
return true;
}
public:
bool isValidBST(TreeNode* root) {
prev = NULL;
return f(root);
}
};
99. Recover Binary Search Tree
- 二叉树搜索树中序遍历得到升序序列
- 用递归不符合题目 O(1) 空间的目标
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
TreeNode* prev = NULL;
TreeNode* broken = NULL;
TreeNode* broken2 = NULL;
bool f(TreeNode* node){
if(!node)
return false;
if(f(node->left))
return true;
if(prev && prev->val>node->val){
if(broken==NULL){
broken = prev;
broken2 = node;
}else{
broken2 = node;
return true;
}
}
prev = node;
if(f(node->right))
return true;
return false;
}
public:
void recoverTree(TreeNode* root) {
f(root);
swap(broken->val, broken2->val);
}
};
100. Same Tree
- 二叉树
- 递归遍历
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
bool isSameTree(struct TreeNode* p, struct TreeNode* q) {
if(p && q)
return (
p->val == q->val &&
isSameTree(p->left, q->left) &&
isSameTree(p->right, q->right)
);
else
return p==NULL && q==NULL;
}
101. Symmetric Tree
- 题目要求两种解法
递归
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
bool isMirror(struct TreeNode* t1, struct TreeNode* t2){
if(!t1 && !t2)
return true;
if(!t1 || !t2)
return false;
if(t1->val!=t2->val)
return false;
return isMirror(t1->left, t2->right) && isMirror(t1->right, t2->left);
}
bool isSymmetric(struct TreeNode* root) {
return root ? isMirror(root->left, root->right) : true;
}
迭代
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isSymmetric(TreeNode* root) {
if(!root)
return true;
stack<TreeNode*> q;
q.push(root->left);
q.push(root->right);
while(!q.empty()){
auto a = q.top();
q.pop();
auto b = q.top();
q.pop();
if(!a && !b)
continue;
if(!a || !b)
return false;
if(a->val!=b->val)
return false;
q.push(a->left);
q.push(b->right);
q.push(b->left);
q.push(a->right);
}
return true;
}
};
这题 LeetCode 目前给的答案不太好
102. Binary Tree Level Order Traversal
- 二叉树层次遍历
- 队列，广度优先
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int>> result;
vector<int> line;
queue<TreeNode*> q;
q.push(root);
q.push(NULL);
for(;;){
TreeNode* node = q.front();
q.pop();
if(node){
line.push_back(node->val);
if(node->left)
q.push(node->left);
if(node->right)
q.push(node->right);
}else{
if(line.empty())
break;
result.push_back(line);
line.clear();
q.push(NULL);
}
}
return result;
}
};
103. Binary Tree Zigzag Level Order Traversal
- 二叉树层次遍历
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
vector<vector<int>> result;
queue<TreeNode*> q;
bool r = false;
if(root)
q.push(root);
while(!q.empty()){
vector<int> item;
int n = q.size();
while(n--){
auto node = q.front();
q.pop();
item.push_back(node->val);
if(node->left)
q.push(node->left);
if(node->right)
q.push(node->right);
}
if(r)
reverse(item.begin(), item.end());
result.push_back(item);
r = !r;
}
return result;
}
};
104. Maximum Depth of Binary Tree
- 二叉树遍历
class Solution {
public:
int maxDepth(TreeNode* root) {
if(root==NULL)
return 0;
return 1 + max(maxDepth(root->left),maxDepth(root->right));
}
};
105. Construct Binary Tree from Preorder and Inorder Traversal
- 二叉树遍历，逆
- 递归
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
function<TreeNode*(int,int,int)> build = [&](int i, int j, int n)->TreeNode*{
if(n==0)
return nullptr;
int val = preorder[i];
int left_size = find(inorder.begin()+j,inorder.end(),val)-inorder.begin()-j;
int right_size = n-left_size-1;
cout<<val<<left_size<<right_size<<endl;
auto node = new TreeNode(val);
node->left = build(i+1, j, left_size);
node->right = build(i+left_size+1, j+left_size+1, right_size);
return node;
};
return build(0,0, preorder.size());
}
};
106. Construct Binary Tree from Inorder and Postorder Traversal
- 同上
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
int find(int *xs, int x){
for(int i=0;;i++)
if(xs[i]==x)
return i;
}
struct TreeNode *build(int *inorder,int *postorder,int n){
if(n==0)
return NULL;
int value = postorder[n-1];
int left_length = find(inorder, value);
int right_length = n - left_length - 1;
struct TreeNode *node = (struct TreeNode*)malloc(sizeof(struct TreeNode));
node->val = value;
node->left = build(inorder, postorder, left_length);
node->right = build(inorder+left_length+1, postorder+left_length, right_length);
return node;
}
struct TreeNode* buildTree(int* inorder, int inorderSize, int* postorder, int postorderSize) {
return build(inorder, postorder, inorderSize);
}
107. Binary Tree Level Order Traversal II
- 二叉树
- 广度优先搜索，层次遍历
- 这题要求输出逆序
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
vector<vector<int>> result;
queue<TreeNode*> q;
if(root)
q.push(root);
while(!q.empty()){
int n = q.size();
vector<int> item;
while(n--){
auto node = q.front();
q.pop();
item.push_back(node->val);
if(node->left)
q.push(node->left);
if(node->right)
q.push(node->right);
}
result.push_back(item);
}
reverse(result.begin(), result.end());
return result;
}
};
108. Convert Sorted Array to Binary Search Tree
- 二分搜索
- 递归，构建搜索树
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
struct TreeNode* sortedArrayToBST(int* nums, int numsSize) {
if(numsSize==0)
return NULL;
typedef struct TreeNode Node;
Node* node = (Node*)malloc(sizeof(Node));
node->val = nums[numsSize/2];
node->left = sortedArrayToBST(nums, numsSize/2);
node->right = sortedArrayToBST(nums+numsSize/2+1, numsSize-numsSize/2-1);
return node;
}
109. Convert Sorted List to Binary Search Tree
- 递归
- 按中序遍历从有序序列中取元素
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
TreeNode* f(ListNode** node, int a, int b){
if(a>=b)
return NULL;
int c = a+(b-a)/2;
auto tree_node = new TreeNode(-1);
tree_node->left = f(node, a, c);
tree_node->val = (*node)->val;
*node = (*node)->next;
tree_node->right = f(node, c+1, b);
return tree_node;
}
public:
TreeNode* sortedListToBST(ListNode* head) {
int n = 0;
for(auto p=head; p; p=p->next)
n++;
return f(&head, 0, n);
}
};
110. Balanced Binary Tree
- 递归+短路
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
bool isBalanced(TreeNode* root, int* depth) {
if(!root){
*depth = 0;
return true;
}
int left_depth, right_depth;
if(!isBalanced(root->left, &left_depth))
return false;
if(!isBalanced(root->right, &right_depth))
return false;
if(abs(left_depth-right_depth)>1)
return false;
*depth = max(left_depth, right_depth)+1;
return true;
}
public:
bool isBalanced(TreeNode* root) {
int depth;
return isBalanced(root, &depth);
}
};
111. Minimum Depth of Binary Tree
- 二叉树
- 递归
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int minDepth(TreeNode* root) {
if(root==nullptr)
return 0;
if(root->left==nullptr)
return minDepth(root->right)+1;
if(root->right==nullptr)
return minDepth(root->left)+1;
return min(minDepth(root->left), minDepth(root->right)) + 1;
}
};
112. Path Sum
- 二叉树，短路
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
bool hasPathSum(struct TreeNode* root, int sum) {
if(!root)
return false;
sum -= root->val;
if(!root->left && !root->right)
return sum==0;
return hasPathSum(root->left, sum) || hasPathSum(root->right, sum);
}
113. Path Sum II
- 二叉树，叶节点，路径
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> pathSum(TreeNode* root, int sum) {
vector<vector<int>> result;
vector<int> path;
function<void(TreeNode*,int)> f = [&](TreeNode* node,int sum){
if(!node)
return;
sum -= node->val;
path.push_back(node->val);
if(!node->left && !node->right){
if(sum==0)
result.push_back(path);
}else{
f(node->left, sum);
f(node->right, sum);
}
path.pop_back();
};
f(root, sum);
return result;
}
};
114. Flatten Binary Tree to Linked List
Tree
- 先序
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
TreeNode* f(TreeNode* node,TreeNode* tail){
if(!node)
return tail;
node->right = f(node->left, f(node->right, tail));
node->left = NULL;
return node;
}
public:
void flatten(TreeNode* root) {
f(root, NULL);
}
};
115. Distinct Subsequences
- 动态规划
- 返回子序列个数
- 递归前缀的部分
class Solution {
public:
int numDistinct(string s, string t) {
vector<int> d(t.size()+1, 0);
d[0] = 1;
for(int i=0; i<s.size(); i++)
for(int j=t.size(); j>0; j--)
if(s[i]==t[j-1])
d[j] += d[j-1];
return d[t.size()];
}
};
116. Populating Next Right Pointers in Each Node
- 这题的输入限定为满二叉树（最后一层也是满的的完全二叉树）。
- 题目要求不开空间，
/**
* Definition for binary tree with next pointer.
* struct TreeLinkNode {
* int val;
* TreeLinkNode *left, *right, *next;
* TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
* };
*/
class Solution {
public:
void connect(TreeLinkNode *root) {
if(!root)
return;
TreeLinkNode* next_level = root;
for(TreeLinkNode* node;;){
node = next_level;
next_level = node->left;
if(!next_level)
break;
TreeLinkNode* prev = NULL;
while(node){
if(prev)
prev->next = node->left;
node->left->next = node->right;
prev = node->right;
node = node->next;
}
}
}
};
117. Populating Next Right Pointers in Each Node II
- 上一题的超集
/**
* Definition for binary tree with next pointer.
* struct TreeLinkNode {
* int val;
* TreeLinkNode *left, *right, *next;
* TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
* };
*/
class Solution {
public:
void connect(TreeLinkNode *root) {
TreeLinkNode *next_level = root;
while(next_level){
auto node = next_level;
next_level = NULL;
TreeLinkNode *prev = NULL;
while(node){
if(node->left){
if(prev)
prev->next = node->left;
if(!next_level)
next_level = node->left;
prev = node->left;
}
if(node->right){
if(prev)
prev->next = node->right;
if(!next_level)
next_level = node->right;
prev = node->right;
}
node = node->next;
}
}
}
};
118. Pascal’s Triangle
class Solution {
public:
vector<vector<int>> generate(int numRows) {
vector<vector<int>> result;
if(numRows==0)
return result;
result.push_back({1});
for(int i=1;i<numRows;i++){
auto& last = result.back();
vector<int> row(i+1);
row[0] = 1;
for(int j=0; j<i; j++)
row[j+1] = last[j] + last[j+1];
row[i] = 1;
result.push_back(row);
}
return result;
}
};
119. Pascal’s Triangle II
class Solution {
public:
vector<int> getRow(int rowIndex) {
vector<int> row(rowIndex+1, 0);
row[0] = 1;
for(int i=0; i<rowIndex; i++)
for(int j=i+1;j>0;j--)
row[j] += row[j-1];
return row;
}
};
120. Triangle
- 动态规划
class Solution {
public:
int minimumTotal(vector<vector<int>>& triangle) {
vector<int> d = triangle.back();
for(int i=triangle.size()-2; i>=0; i--){
for(int j=0; j<i+1; j++){
d[j] = triangle[i][j] + min(d[j],d[j+1]);
}
}
return d[0];
}
};
121. Best Time to Buy and Sell Stock
- 一次买，一次卖
- 先买后卖，低买高卖
- 遍历序列，记录当前的最小值
class Solution {
public:
int maxProfit(vector<int>& prices) {
if(prices.size()<=1)
return 0;
int lowest = prices[0], profit = 0;
for(int x: prices){
lowest = min(lowest, x);
profit = max(x-lowest, profit);
}
return profit;
}
};
class Solution {
public:
int maxProfit(vector<int>& prices) {
if(prices.size()<=1)
return 0;
int lowest = INT_MAX, profit = 0;
for(int x: prices){
lowest = min(lowest, x);
profit = max(x-lowest, profit);
}
return profit;
}
};
122. Best Time to Buy and Sell Stock II
- 要求买入前已经卖出
class Solution {
public:
int maxProfit(vector<int>& prices) {
int profit = 0;
for(int i=1; i<prices.size(); i++)
profit += max(prices[i]-prices[i-1], 0);
return profit;
}
};
123. Best Time to Buy and Sell Stock III
- 最多两次交易
- 前后扫两趟，分治
class Solution {
public:
int maxProfit(vector<int>& prices) {
int n = prices.size();
if(n<=1)
return 0;
vector<int> profit1(n, 0);
vector<int> profit2(n, 0);
int lowest = prices[0];
for(int i=1; i<n; i++){
lowest = min(lowest, prices[i]);
profit1[i] = max(prices[i]-lowest, profit1[i-1]);
}
cout<<profit1[n-1]<<lowest<<endl;
int highest = prices[n-1];
for(int i=n-2; i>=0; i--){
highest = max(highest, prices[i]);
profit2[i] = max(highest-prices[i], profit2[i+1]);
}
cout<<profit2[0]<<highest<<endl;
int profit = INT_MIN;
for(int i=0; i<n; i++){
profit = max(profit, profit1[i]+profit2[i]);
}
return profit;
}
};
124. Binary Tree Maximum Path Sum
- 加正数会变大
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
int ans = INT_MIN;
int f(TreeNode* node){
if(node){
int val = node->val;
int left = max(f(node->left), 0);
int right = max(f(node->right), 0);
ans = max(ans, left+right+val);
return max(left,right)+val;
}else{
return 0;
}
}
public:
int maxPathSum(TreeNode* root) {
f(root);
return ans;
}
};
125. Valid Palindrome
class Solution {
public:
bool isPalindrome(string s) {
if(s.empty())
return true;
for(int i=0, j=s.size()-1;;i++,j--){
while(!isalnum(s[i]))
i++;
while(!isalnum(s[j]))
j--;
if(i>=j)
break;
if(toupper(s[i])!=toupper(s[j]))
return false;
}
return true;
}
};
126. Word Ladder II
- 图
class Solution {
template<class Callback>
inline void transform(string word,Callback callback){
for(int i=0; i<word.size(); i++){
char c = word[i];
for(char j='a';j<='z';j++){
if(j!=c){
word[i] = j;
callback(word);
}
}
word[i] = c;
}
}
public:
vector<vector<string>> findLadders(string beginWord, string endWord, unordered_set<string> &wordList) {
unordered_map<string,int> dist;
unordered_map<string,vector<string>> path;
queue<string> q;
q.push(beginWord);
dist[beginWord] = 1;
while(!q.empty()){
auto x = q.front();
q.pop();
cout<<x<<endl;
if(x==endWord)
break;
transform(x, [&](const string& word){
if(wordList.count(word)){
if(dist[word]==0){
dist[word] = dist[x]+1;
q.push(word);
}
if(dist[word]==dist[x]+1)
path[word].push_back(x);
}
});
}
vector<vector<string>> result;
vector<string> item;
function<void(const string&)> buildPath = [&](const string& node){
if(node==beginWord){
auto item2 = item;
item2.push_back(node);
reverse(begin(item2),end(item2));
result.push_back(item2);
return;
}
item.push_back(node);
for(auto& s:path[node])
buildPath(s);
item.pop_back();
};
buildPath(endWord);
return result;
}
};
127. Word Ladder
- 图论，广度优先搜索
- 用队列实现
- 这题给个测试用例单词很短，但是单词数量很多
- 比如说有个 5 的字符长度（5*25=125个变换），2370 个单词的
- 这种情况下用字符串两两比较会超时
class Solution {
bool adjacent(const string& s1, const string& s2){
if(s1.size()!=s2.size())
return false;
int count = 0;
for(int i=0; i<s1.size() && count<2; i++){
if(s1[i]!=s2[i])
count++;
}
return count==1;
}
public:
int ladderLength(string beginWord, string endWord, unordered_set<string>& wordList) {
unordered_map<string,int> dist;
queue<string> q;
q.push(beginWord);
dist[beginWord] = 1;
while(!q.empty()){
auto x = q.front();
q.pop();
if(x==endWord)
break;
string o = x;
for(int i=0; i<x.size(); i++){
char c = x[i];
for(char j='a';j<='z';j++){
if(j!=c){
x[i] = j;
if(wordList.count(x) && !dist[x]){
q.push(x);
dist[x] = dist[o]+1;
}
}
}
x[i] = c;
}
// for(auto v: wordList){
// if(!dist[v] && adjacent(x,v)){
// q.push(v);
// dist[v] = dist[x]+1;
// }
// }
}
return dist[endWord];
}
};
128. Longest Consecutive Sequence
- 要求 O(N)
class Solution {
unordered_map<int,int> h;
int ds_find(int x){
if(h[x]!=x)
h[x] = ds_find(h[x]);
return h[x];
}
void ds_union(int x, int y){
h[ds_find(x)] = ds_find(y);
}
public:
int longestConsecutive(vector<int>& nums) {
for(int x: nums)
h[x] = x;
for(int x: nums){
if(h.count(x+1))
ds_union(x, x+1);
if(h.count(x-1))
ds_union(x, x-1);
}
unordered_map<int,int> a;
for(auto p: h)
a[ds_find(p.second)]++;
int ans = 0;
for(auto p: a)
ans = max(ans, p.second);
return ans;
}
};
129. Sum Root to Leaf Numbers
- 二叉树
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int sumNumbers(TreeNode* root) {
function<int(TreeNode*,int)> f = [&](TreeNode* node,int num){
if(!node)
return 0;
num = num*10 + node->val;
if(!node->left && !node->right)
return num;
return f(node->left, num)+f(node->right, num);
};
return f(root, 0);
}
};
130. Surrounded Regions
- 递归会爆栈，
- LeetCode 显示 RunTime Error
class Solution {
public:
void solve(vector<vector<char>>& board) {
int m = board.size();
if(m==0)
return;
int n = board[0].size();
function<void(int,int)> f = [&](int x, int y){
if(board[y][x]=='O'){
board[y][x] = 'T';
if(y+1 < m-1)
f(x, y+1);
if(y-1 > 0)
f(x, y-1);
if(x+1 < n-1)
f(x+1, y);
if(x-1 > 0)
f(x-1, y);
}
};
for(int i=0; i<n; i++){
f(i, 0);
f(i, m-1);
}
for(int j=0; j<m; j++){
f(0, j);
f(n-1, j);
}
for(auto& xs: board)
for(auto& x: xs)
if(x=='O')
x = 'X';
else if(x=='T')
x = 'O';
}
};
131. Palindrome Partitioning
Backtracking
- 深度优先搜索
- 动态规划
class Solution {
bool isPalindrome(const string& s, int start, int last){
while(start<last){
if(s[start]!=s[last])
return false;
start++, last--;
}
return true;
}
public:
vector<vector<string>> partition(string s) {
vector<vector<string>> result;
//unordered_map<string,vector<string>> h;
vector<string> path;
function<void(int)> f = [&](int i){
if(i==s.size()){
result.push_back(path);
return;
}
for(int j=i; j<s.size(); j++){
if(isPalindrome(s,i,j)){
path.push_back(s.substr(i,j-i+1));
f(j+1);
path.pop_back();
}
}
};
f(0);
return result;
}
};
132. Palindrome Partitioning II
- 求最小割数
- 长度 n 最多 n-1 个割
- 一个字串可以递归分解为一个回文子串和另一个字串
class Solution {
public:
int minCut(string s) {
int n = s.size();
vector<int> c(n+1);
vector<vector<bool>> d(n,vector<bool>(n));
for(int i=0; i<=n; i++)
c[i] = i - 1;
for(int i=0; i<n; i++)
for(int j=i; j>=0; j--)
if(s[j]==s[i] && (i-j<3 || d[j+1][i-1])){
d[j][i] = true;
c[i+1] = min(c[i+1], c[j]+1);
}
return c[n];
}
};
class Solution {
public:
int minCut(string s) {
int n = s.size();
vector<int> c(n+1);
vector<vector<bool>> d(n,vector<bool>(n));
for(int i=0; i<=n; i++)
c[i] = i - 1;
for(int i=0; i<n; i++)
for(int j=0; j<=i; j++)
if(s[j]==s[i] && (i-j<3 || d[j+1][i-1])){
d[j][i] = true;
c[i+1] = min(c[i+1], c[j]+1);
}
return c[n];
}
};
133. Clone Graph
- 图
- 哈希表
- 递归
/**
* Definition for undirected graph.
* struct UndirectedGraphNode {
* int label;
* vector<UndirectedGraphNode *> neighbors;
* UndirectedGraphNode(int x) : label(x) {};
* };
*/
class Solution {
public:
UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
unordered_map<UndirectedGraphNode*,UndirectedGraphNode*> h;
function<UndirectedGraphNode*(UndirectedGraphNode*)> f = [&](UndirectedGraphNode* node){
auto it = h.find(node);
if(it!=h.end())
return it->second;
auto new_node = new UndirectedGraphNode(node->label);
h[node] = new_node;
for(auto e: node->neighbors)
new_node->neighbors.push_back(f(e));
return new_node;
};
return node ? f(node) : nullptr;
}
};
134. Gas Station
- 一趟
- 有两个目标
- 判断是否能完成
- 寻找起点
class Solution {
public:
int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
int a = 0, d = 0;
int start = 0;
for(int i=0; i<gas.size(); i++){
d += gas[i] - cost[i];
if(d<0){
a += d;
d = 0;
start = i+1;
}
}
return a+d>=0 ? start : -1;
}
};
135. Candy
- 记录变化量
int candy(int* ratings, int ratingsSize) {
int a, *d = (int*)calloc(ratingsSize,sizeof(int));
a = 1;
for(int i=1; i<ratingsSize; i++){
if(ratings[i] > ratings[i-1]){
if(d[i]<a)
d[i] = a;
a++;
}else{
a = 1;
}
}
a = 1;
for(int i=ratingsSize-2; i>=0; i--){
if(ratings[i] > ratings[i+1]){
if(d[i]<a)
d[i] = a;
a++;
}else{
a = 1;
}
}
int sum = ratingsSize;
for(int i=0; i<ratingsSize; i++){
sum += d[i];
}
return sum;
}
136. Single Number
- 整数的位运算
- 异或运算，和零异或为自己，和自己异或为 0
int singleNumber(int* nums, int numsSize) {
int x = 0;
for(int i=0; i<numsSize; i++)
x ^= nums[i];
return x;
}
137. Single Number II
- 二进制位加法
- 异或运算得到值，与运算得到进位
- 本题得到 0b11 时再取模
class Solution {
public:
int singleNumber(vector<int>& nums) {
int a=0, b=0;
for(int x: nums){
b ^= a&x;
a ^= x;
int c = ~(a&b);
a &= c;
b &= c;
}
return a;
}
};
138. Copy List with Random Pointer
- 图的复制
/**
* Definition for singly-linked list with a random pointer.
* struct RandomListNode {
* int label;
* struct RandomListNode *next;
* struct RandomListNode *random;
* };
*/
struct RandomListNode *copyRandomList(struct RandomListNode *head) {
typedef struct RandomListNode Node;
if(!head)
return NULL;
for(Node* node=head; node;){
Node* copy = (Node*)malloc(sizeof(Node));
*copy = *node;
node->next = copy;
node = copy->next;
}
for(Node* node=head; node;){
Node* copy = node->next;
if(copy->random)
copy->random = copy->random->next;
node = copy->next;
}
Node *head2 = head->next;
for(Node *node = head; node;){
Node *node2 = node->next;
Node *next = node2->next;
node->next = next;
node2->next = next ? next->next : NULL;
node = next;
}
return head2;
}
/**
* Definition for singly-linked list with a random pointer.
* struct RandomListNode {
* int label;
* RandomListNode *next, *random;
* RandomListNode(int x) : label(x), next(NULL), random(NULL) {}
* };
*/
class Solution {
public:
RandomListNode *copyRandomList(RandomListNode *head) {
unordered_map<RandomListNode*,RandomListNode*> h;
auto get = [&](RandomListNode* node)->RandomListNode*{
if(!node)
return NULL;
auto it = h.find(node);
if(it!=h.end())
return it->second;
else{
return h[node] = new RandomListNode(node->label);
}
};
for(auto p=head; p; p=p->next){
auto node_clone = get(p);
node_clone->random = get(p->random);
node_clone->next = get(p->next);
}
return h[head];
}
};
139. Word Break
- 动态规划
- 分词/递归
class Solution {
public:
bool wordBreak(string s, vector<string>& wordDict) {
unordered_set<string> dict;
for(auto& x: wordDict)
dict.insert(x);
vector<bool> d(s.size()+1, false);
d[0] = true;
for(int i=1;i<=s.size();i++)
for(int j=i-1;j>=0;j--)
if(d[j] && dict.count(s.substr(j, i-j))){
d[i] = true;
break;
}
return d.back();
}
};
class Solution {
public:
bool wordBreak(string s, vector<string>& wordDict) {
unordered_set<string> dict;
int len = 0;
for(auto& x: wordDict){
dict.insert(x);
len = max(len,(int)x.size());
}
unordered_map<int,bool> m;
function<bool(int)> f = [&](int i){
if(i==s.size())
return true;
auto it = m.find(i);
if(it!=m.end())
return it->second;
for(int j=i+1; j<=s.size() && j-i<=len; j++)
if(dict.count(s.substr(i,j-i)) && f(j))
return m[i] = true;
return m[i] = false;
};
return f(0);
}
};
140. Word Break II
class Solution {
public:
vector<string> wordBreak(string s, vector<string>& wordDict) {
unordered_set<string> dict;
for(auto& x: wordDict)
dict.insert(x);
int n = s.size();
vector<vector<int>> edge(n+1,vector<int>());
edge[0].push_back(0);
for(int i=1; i<=n; i++)
for(int j=i-1; j>=0; j--)
if(edge[j].size() && dict.count(s.substr(j, i-j))){
edge[i].push_back(j);
printf("(%d->%d)\n",j,i);
}
vector<string> result;
vector<string> path;
function<void(int)> f = [&](int i){
if(i==0){
string item;
for(int j=path.size()-1; j>=0; j--){
item += path[j];
item += ' ';
}
item.pop_back();
result.push_back(item);
return;
}
for(auto j: edge[i]){
path.push_back(s.substr(j, i-j));
f(j);
path.pop_back();
}
};
f(n);
return result;
}
};
141. Linked List Cycle
- 链表，双指针
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
bool hasCycle(ListNode *head) {
auto fast = head, slow = head;
while(fast && fast->next){
fast = fast->next->next;
slow = slow->next;
if(fast==slow)
return true;
}
return false;
}
};
142. Linked List Cycle II
List
- 链表
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *detectCycle(ListNode *head) {
auto fast = head, slow = head;
while(fast && fast->next){
fast = fast->next->next;
slow = slow->next;
if(fast==slow){
auto node = head;
while(node!=slow){
slow = slow->next;
node = node->next;
}
return node;
}
}
return NULL;
}
};
143. Reorder List
- 链表
- 原地逆置+链表合并+快慢双指针
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
ListNode* nreverse(ListNode* head){
ListNode* prev = NULL;
while(head){
auto next = head->next;
head->next = prev;
prev = head;
head = next;
}
return prev;
}
void merge(ListNode* head1,ListNode* head2){
while(head2){
auto next1 = head1->next;
auto next2 = head2->next;
head1->next = head2;
head2->next = next1;
head1 = next1;
head2 = next2;
}
}
ListNode* cut(ListNode* head){
auto fast = head, slow = head;
while(fast && fast->next){
fast = fast->next->next;
slow = slow->next;
}
auto head2 = slow->next;
slow->next = NULL;
return head2;
}
public:
void reorderList(ListNode* head) {
if(head)
merge(head,nreverse(cut(head)));
}
};
144. Binary Tree Preorder Traversal
- 重点问题！
- 二叉树先序遍历
- NLR，先访问根节点
- 递归，非递归实现
- 出栈顺序
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> preorderTraversal(TreeNode* root) {
vector<int> result;
stack<TreeNode*> s;
if(root)
s.push(root);
while(!s.empty()){
TreeNode* node = s.top();
s.pop();
result.push_back(node->val);
if(node->right)
s.push(node->right);
if(node->left)
s.push(node->left);
}
return result;
}
};
145. Binary Tree Postorder Traversal
Tree
- 二叉树后序遍历的迭代实现
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> postorderTraversal(TreeNode* root) {
vector<int> result;
stack<TreeNode*> s;
TreeNode *p = root;
TreeNode *q = NULL;
for(;;){
if(p){
s.push(p);
p = p->left;
}else if(!s.empty()){
p = s.top();
if(q==p->right || !p->right){
result.push_back(p->val);
q = p;
s.pop();
p = NULL;
}else{
p = p->right;
}
}else{
break;
}
}
return result;
}
};
146. LRU Cache
- 需要在 O(1) 时间找到链表中的节点并移动到头部
- 利用双向链表和哈希表
- c++ 的 splice 用来把另一个链表中的元素插入该位置前
class LRUCache{
int capacity;
unordered_map<int,list<pair<int,int>>::iterator> h;
list<pair<int,int>> l;
public:
LRUCache(int capacity) {
this->capacity = capacity;
}
int get(int key) {
auto it = h.find(key);
if(it==h.end())
return -1;
l.splice(l.begin(), l, it->second);
return it->second->second;
}
void set(int key, int value) {
auto it = h.find(key);
if(it!=h.end()){
it->second->second = value;
l.splice(l.begin(), l, it->second);
}else{
if(l.size()==capacity){
h.erase(l.back().first);
l.pop_back();
}
l.push_front(make_pair(key, value));
h[key] = l.begin();
}
}
};
147. Insertion Sort List
- 数组一般从后往前插
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* insertionSortList(ListNode* head) {
ListNode* list = NULL;
auto node = head;
while(node){
auto next = node->next;
auto p = &list;
while(*p && node->val > (*p)->val)
p = &(*p)->next;
node->next = *p;
*p = node;
node = next;
}
return list;
}
};
148. Sort List
- 链表
- 归并排序
- 结合前面的题目
-
21. Merge Two Sorted Lists，直接搬过来递归使用 -
143. Reorder List，那题有两个节点的话不用断开，这题要。
-
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
ListNode* cut(ListNode* head){
auto fast = head, slow = head;
while(fast && fast->next && fast->next->next){
fast = fast->next->next;
slow = slow->next;
}
auto head2 = slow->next;
slow->next = NULL;
return head2;
}
ListNode* nmerge(ListNode* head1,ListNode* head2){
ListNode* head = NULL;
ListNode** append = &head;
for(;;){
if(!head1){
*append = head2;
break;
}
if(!head2){
*append = head1;
break;
}
if(head1->val < head2->val){
*append = head1;
head1 = head1->next;
}else{
*append = head2;
head2 = head2->next;
}
append = &(*append)->next;
}
return head;
}
public:
ListNode* sortList(ListNode* head) {
if(!head || !head->next)
return head;
auto head2 = cut(head);
return nmerge(sortList(head),sortList(head2));
}
};
149. Max Points on a Line
- 注意出现重合的点的情况
- [[1,1],[1,1],[1,1]]
- 该题的坐标是整数
/**
* Definition for a point.
* struct Point {
* int x;
* int y;
* Point() : x(0), y(0) {}
* Point(int a, int b) : x(a), y(b) {}
* };
*/
class Solution {
public:
int maxPoints(vector<Point>& points) {
int n = points.size();
if(n<=2)
return points.size();
int m = 2;
for(int i=0; i<n; i++){
int start = 2;
for(int j=i+1; j<n-1; j++){
int dx = points[j].x-points[i].x;
int dy = points[j].y-points[i].y;
if(dx==0 && dy==0){
start++;
continue;
}
int bdx = dx*points[i].y - dy*points[i].x;
int count = start;
for(int k=j+1; k<n; k++){
if(dx==0){
if(points[k].x == points[j].x)
count++;
}else{
if(dx*points[k].y==dy*points[k].x+bdx)
count++;
}
}
m = max(m, count);
}
m = max(m, start);
}
return m;
}
};
150. Evaluate Reverse Polish Notation
- Stack
int evalRPN(char** tokens, int tokensSize) {
int size = 3;
int* stack = (int*)malloc(sizeof(int)*size);
int *p = stack;
for(int i=0; i<tokensSize; i++){
char *token = tokens[i];
if(token[0] && !token[1]){
switch(token[0]){
case '+':
p[-2] += p[-1];
p--;
continue;
case '-':
p[-2] -= p[-1];
p--;
continue;
case '*':
p[-2] *= p[-1];
p--;
continue;
case '/':
p[-2] /= p[-1];
p--;
continue;
}
}
if(p-stack==size){
stack = realloc(stack,sizeof(int)*(size*2));
p = stack+size;
size *= 2;
}
*p++ = atoi(token);
}
int ans = p[-1];
free(stack);
return ans;
}
151. Reverse Words in a String
Application
- 逆置
- 原地，一趟。
class Solution {
void reverse(string& s, int i, int j){
while(i<j)
swap(s[i++],s[j--]);
}
void compact(string& s){
int k = 0;
for(int i=0; i<s.size(); i++)
if(s[i]!=' '){
if(k!=0)
s[k++] = ' ';
int j = i;
while(j<s.size() && s[j]!=' ')
s[k++] = s[j++];
i = j;
}
s.resize(k);
}
public:
void reverseWords(string &s) {
compact(s);
::reverse(s.begin(), s.end());
int start = 0;
while(start<s.size() && s[start]==' ')
start++;
for(int i=0; i<s.size(); i++){
if(s[i]==' '){
reverse(s, start, i-1);
start = i+1;
}
}
reverse(s, start, s.size()-1);
}
};
152. Maximum Product Subarray
- 积最大
- 正负
class Solution {
public:
int maxProduct(vector<int>& nums) {
int m = INT_MIN;
int a = 1, b = 1;
for(int x: nums){
int c = max(x, max(a*x, b*x));
b = min(x, min(a*x, b*x));
m = max(m, c);
a = c;
}
return m;
}
};
153. Find Minimum in Rotated Sorted Array
这个解答有问题。。。
class Solution {
public:
int findMin(vector<int>& nums) {
int a = 0, b = nums.size();
while(a<b){
int c = (a+b-1)/2;
if(nums[a]<=nums[b-1]){
return nums[a];
}else if(nums[c]>nums[b-1]){
a = c + 1;
}else{
b = c + 1;
}
}
return INT_MIN;
}
};
154. Find Minimum in Rotated Sorted Array II
- 有重复元素
class Solution {
public:
int findMin(vector<int>& nums) {
int a = 0, b = nums.size()-1;
while(a<b){
int c = (a+b)/2;
if(nums[a]<nums[b])
break;
else if(nums[c]>nums[b])
a = c + 1;
else if(nums[a]>nums[c])
b = c;
else
a++;
}
return nums[a];
}
};
class Solution {
public:
int findMin(vector<int>& nums) {
int a = 0, b = nums.size();
while(b-a>1){
int c = (a+b-1)/2;
if(nums[a]<nums[b-1])
break;
else if(nums[c]>nums[b-1])
a = c + 1;
else if(nums[a]>nums[c])
b = c + 1;
else
a++;
}
return nums[a];
}
};
155. Min Stack
- 栈，先进先出
- 用另一个栈记录最小元素
- STL 中空栈取元素是未定义行为
class MinStack {
stack<int> _stack;
stack<int> _stack_min;
public:
/** initialize your data structure here. */
MinStack() {
}
void push(int x) {
_stack.push(x);
if(_stack_min.empty() || x<=_stack_min.top())
_stack_min.push(x);
}
void pop() {
int x = _stack.top();
_stack.pop();
if(!_stack_min.empty() && x==_stack_min.top())
_stack_min.pop();
}
int top() {
return _stack.top();
}
int getMin() {
return _stack_min.top();
}
};
/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(x);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.getMin();
*/
160. Intersection of Two Linked Lists
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
int list_length(ListNode* head){
int n=1;
while(head){
n++;
head = head->next;
}
return n;
}
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
int lenA = list_length(headA);
int lenB = list_length(headB);
if(lenA>lenB){
for(int i=lenA-lenB; i>0; i--)
headA = headA->next;
}else{
for(int i=lenB-lenA; i>0; i--)
headB = headB->next;
}
while(headA!=headB){
headA = headA->next;
headB = headB->next;
}
return headA;
}
};
162. Find Peak Element
- 二分查找
- 找极大值
- 越界为负无穷
class Solution {
public:
int findPeakElement(vector<int>& nums) {
auto less = [&](int i, int j){
if(i==-1 || i==nums.size())
return true;
if(j==-1 || j==nums.size())
return false;
return nums[i] < nums[j];
};
int a = 0, b = nums.size();
while(a<b){
int c = (a+b)/2;
if(less(c-1,c) && less(c, c+1))
a = c + 1;
else if(less(c+1, c) && less(c, c-1))
b = c;
else if(less(c+1, c) && less(c-1, c))
return c;
else
b = c;
}
return -1;
}
};
164. Maximum Gap
Sort
- 先排序 O(N)，后查找 O(N)
//http://zh.wikipedia.org/zh-cn/%E5%9F%BA%E6%95%B0%E6%8E%92%E5%BA%8F
void radixsort(int data[], int n){
int m = 0;
for(int i=0;i<n;i++)
if(data[i]>m)
m = data[i];
int *tmp = (int*)malloc(sizeof(int)*n);
memset(tmp, 0, sizeof(int)*n);
int count[10] = {0};
for(unsigned radix=1;radix<=m;radix*=10){
for(int i=0;i<10;i++)
count[i] = 0;
for(int i=0; i<n; i++)
count[(data[i]/radix)%10]++;
for(int i=1; i<10; i++)
count[i] += count[i-1];
for(int i=n-1; i>=0; i--)
tmp[--count[(data[i]/radix)%10]] = data[i];
for(int i=0; i<n; i++)
data[i] = tmp[i];
}
free(tmp);
}
int maximumGap(int num[], int n) {
if(n<2)
return 0;
radixsort(num,n);
int pred = num[0];
int max = 0;
for(int i=1;i<n;i++){
int curr = num[i];
int delta = curr>pred ? curr-pred : pred-curr;
if(delta>max)
max = delta;
pred = curr;
}
return max;
}
165. Compare Version Numbers
int compareVersion(char* version1, char* version2) {
while(*version1 || *version2){
int a = strtol(version1,&version1,10);
int b = strtol(version2,&version2,10);
if(*version1=='.')version1++;
if(*version2=='.')version2++;
if(a<b)return -1;
if(a>b)return 1;
}
return 0;
}
166. Fraction to Recurring Decimal
class Solution {
void append(string& s, unsigned n){
stringstream ss;
ss << n;
s += ss.str();
}
public:
string fractionToDecimal(int numerator, int denominator) {
string s;
if(numerator==0)
return "0";
if((numerator^denominator)&(1<<31))
s += "-";
long long a = llabs((long long)numerator), b = llabs((long long)denominator);
append(s, a/b);
a %= b;
if(a!=0){
s.push_back('.');
unordered_map<unsigned,int> h;
for(int i=s.size(); a; i++){
if(h.count(a)){
s.insert(h[a], "(");
s.push_back(')');
break;
}
h[a] = i;
a *= 10;
append(s, a/b);
a %= b;
}
}
return s;
}
};
167. Two Sum II – Input array is sorted
- 这题没意思
class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
int i = 0, j = numbers.size()-1;
while(i<j){
int s = numbers[i] + numbers[j];
if(s==target)
return {i+1, j+1};
else if(s<target)
i++;
else
j--;
}
return {0,0};
}
};
168. Excel Sheet Column Title
class Solution {
public:
string convertToTitle(int n) {
string s;
while(n){
n--;
s += 'A'+ n%26;
n /= 26;
}
reverse(s.begin(), s.end());
return s;
}
};
169. Majority Element
- 数组老题，题干有若干假设
int majorityElement(int* nums, int numsSize) {
int x = nums[0];
int count = 1;
for(int i=1; i<numsSize; i++){
if(nums[i]==x){
count++;
}else{
count--;
if(count==0){
x = nums[i];
count=1;
}
}
}
return x;
}
171. Excel Sheet Column Number
int titleToNumber(char* s) {
int n = 0;
for(;*s;s++)
n = n*26+*s-'A'+1;
return n;
}
172. Factorial Trailing Zeroes
索引：整数
- 求区间上因子5的个数
int trailingZeroes(int n) {
int a = 0;
while(n>=5){
n /= 5;
a += n;
}
return a;
}
173. Binary Search Tree Iterator
- 二叉树的中序遍历
- 写成外迭代器的形式
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class BSTIterator {
stack<TreeNode*> s;
public:
BSTIterator(TreeNode *root) {
auto p = root;
while(p){
s.push(p);
p = p->left;
}
}
/** @return whether we have a next smallest number */
bool hasNext() {
return !s.empty();
}
/** @return the next smallest number */
int next() {
auto p = s.top();
s.pop();
int val = p->val;
p = p->right;
while(p){
s.push(p);
p = p->left;
}
return val;
}
};
/**
* Your BSTIterator will be called like this:
* BSTIterator i = BSTIterator(root);
* while (i.hasNext()) cout << i.next();
*/
174. Dungeon Game
DP
- 深搜，符合DP的两个条件
- 路径上一直大于零。
class Solution {
public:
int calculateMinimumHP(vector<vector<int>>& dungeon) {
int m = dungeon.size();
int n = dungeon[0].size();
int d[n];
d[n-1] = max(-dungeon[m-1][n-1], 0);
for(int j=n-2;j>=0;j--)
d[j] = max(d[j+1]-dungeon[m-1][j], 0);
for(int i=m-2;i>=0;i--){
d[n-1] = max(d[n-1]-dungeon[i][n-1], 0);
for(int j=n-2;j>=0;j--){
int cost = dungeon[i][j];
d[j] = min(d[j]-cost, d[j+1]-cost);
d[j] = max(d[j], 0);
}
}
return d[0]+1;
}
};
179. Largest Number
class Solution {
public:
string largestNumber(vector<int>& nums) {
vector<string> xs;
for(int x: nums){
stringstream ss;
ss << x;
xs.push_back(ss.str());
}
sort(begin(xs), end(xs),[](const string& a, const string& b){
return a+b > b+a;
});
if(xs.size()==0 || xs[0]=="0")
return "0";
stringstream ss;
for(auto& x: xs)
ss << x;
return ss.str();
}
};
187. Repeated DNA Sequences
- 查找
class Solution {
public:
vector<string> findRepeatedDnaSequences(string s) {
vector<string> result;
unordered_map<char, int> m = {{'A',0},{'C',1},{'G',2},{'T',3}};
unordered_map<int, int> h;
int key = 0;
for(int i=0; i<9; i++)
key = (key<<2) | m[s[i]];
for(int i=9; i<s.size(); i++){
key = ((key<<2) | m[s[i]]) & ((1<<20) - 1);
if(++h[key]==2)
result.push_back(s.substr(i-9, 10));
}
return result;
}
};
188. Best Time to Buy and Sell Stock IV
- 买卖交替
- k 很大的情况
- k==2 的情况
class Solution {
public:
int maxProfit(int k, vector<int>& prices) {
if(k>=prices.size()/2){
int ans = 0;
for(int i=1; i<prices.size(); i++){
int d = prices[i] - prices[i-1];
ans += max(d, 0);
}
return ans;
}
vector<int> l(k+1);
vector<int> g(k+1);
for(int i=1; i<prices.size();i++){
int d = prices[i] - prices[i-1];
for(int j=k; j>=1; j--){
l[j] = max(g[j-1]+max(d,0), l[j]+d);
g[j] = max(g[j], l[j]);
}
}
return g[k];
}
};
189. Rotate Array
Array
- 数组问题
- 三次逆置
void reverse(int* nums, int i, int j){
for(; i<j; i++,j--){
int t = nums[i];
nums[i] = nums[j];
nums[j] = t;
}
}
void rotate(int* nums, int numsSize, int k) {
k %= numsSize;
if(k==0)
return;
reverse(nums,0,numsSize-k-1);
reverse(nums,numsSize-k,numsSize-1);
reverse(nums,0,numsSize-1);
}
190. Reverse Bits
uint32_t reverseBits(uint32_t n) {
int a = 0;
for(int i=0; i<32; i++){
a <<= 1;
a |= n&1;
n >>= 1;
}
return a;
}
191. Number of 1 Bits
- 整数位运算经典题
- 一次迭代去掉一个最低位的1
- 发生二进制减法借位
int hammingWeight(uint32_t n) {
int count = 0;
while(n){
count++;
n &= n-1;
}
return count;
}
198. House Robber
DP
- 应用题，动态规划
class Solution {
public:
int rob(vector<int>& nums) {
int n = nums.size();
if(n==0)
return 0;
if(n==1)
return nums[0];
vector<int> d(n);
d[0] = nums[0];
d[1] = max(d[0], nums[1]);
for(int i=2; i<nums.size(); i++)
d[i] = max(d[i-1], d[i-2]+nums[i]);
return d[n-1];
}
};
199. Binary Tree Right Side View
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> rightSideView(TreeNode* root) {
vector<int> result;
queue<TreeNode*> q;
if(root)
q.push(root);
while(!q.empty()){
result.push_back(q.back()->val);
int n = q.size();
while(n--){
auto node = q.front();
q.pop();
if(node->left)
q.push(node->left);
if(node->right)
q.push(node->right);
}
}
return result;
}
};
200. Number of Islands
Graph
void fill(char **grid, int m, int n, int y, int x) {
if(grid[y][x]=='1'){
grid[y][x] = '0';
if(y>0)
fill(grid, m, n, y-1, x);
if(y+1<m)
fill(grid, m, n, y+1, x);
if(x>0)
fill(grid, m, n, y, x-1);
if(x+1<n)
fill(grid, m, n, y, x+1);
}
}
int numIslands(char** grid, int gridRowSize, int gridColSize) {
int count = 0;
for(int y=0;y<gridRowSize;y++)
for(int x=0;x<gridColSize;x++){
if(grid[y][x]=='1'){
fill(grid, gridRowSize, gridColSize, y, x);
count++;
}
}
return count;
}
201. Bitwise AND of Numbers Range
Integer
- 位运算的转化，覆盖区间上的数
class Solution {
public:
int rangeBitwiseAnd(int m, int n) {
unsigned b = -1;
while((m&b)!=(n&b))
b <<= 1;
return m&b;
}
};
202. Happy Number
class Solution {
int next(int x){
int y = 0;
while(x){
y += (x%10)*(x%10);
x /= 10;
}
return y;
}
public:
bool isHappy(int n) {
unordered_map<int,bool> h;
while(n!=1){
if(h.count(n))
return false;
h[n] = true;
n = next(n);
}
return true;
}
};
203. Remove Linked List Elements
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* removeElements(ListNode* head, int val) {
auto list = &head;
while(*list){
if((*list)->val==val)
*list = (*list)->next;
else
list = &(*list)->next;
}
return head;
}
};
204. Count Primes
- 质数
- 筛
int countPrimes(int n) {
int count = 0;
bool h[n];
memset(h,-1,sizeof(bool)*n);
h[0] = h[1] = 0;
for(int i=2; i<n; i++)
if(h[i])
for(int j=i<<1; j<n; j+=i)
h[j] = 0;
for(int i=0; i<n; i++)
if(h[i])
count++;
return count;
}
205. Isomorphic Strings
bool isIsomorphic(char* s, char* t) {
char h[128] = {0}, r[128] = {0};
while(*s){
if(h[*s] == 0 && r[*t]==0){
h[*s] = *t;
r[*t] = *s;
}else if(h[*s]!=*t)
return false;
s++, t++;
}
return true;
}
206. Reverse Linked List
- 链表经典问题
- 原地逆置链表
class Solution {
public:
ListNode* reverseList(ListNode* head) {
ListNode* prev = NULL;
while(head){
ListNode* next = head->next;
head->next = prev;
prev = head;
head = next;
}
return prev;
}
};
207. Course Schedule
Graph
- 有趣的题目
- 有向图拓扑排序，判断是否有环
- 用深度优先搜索实现
class Solution {
public:
bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
unordered_map<int,int> h;
function<bool(int)> f = [&](int x){
if(h[x]==1)
return true;
if(h[x]==2)
return false;
h[x] = 1;
for(auto p: prerequisites)
if(p.first==x){
if(f(p.second))
return true;
}
h[x] = 2;
return false;
};
for(int i=0; i<numCourses; i++)
if(f(i))
return false;
return true;
}
};
208. Implement Trie (Prefix Tree)
- 字符串查找，前缀
- 后面有个题目会用到这个数据结构
class Trie {
struct Node{
Node* next[26];
bool term;
Node(){
memset(this,0,sizeof(Node));
}
};
Node* root;
public:
/** Initialize your data structure here. */
Trie() {
root = new Node();
}
/** Inserts a word into the trie. */
void insert(string word) {
Node* node = root;
for(char c: word){
Node*& next = node->next[c-'a'];
if(!next)
next = new Node();
node = next;
}
node->term = true;
}
/** Returns if the word is in the trie. */
bool search(string word) {
Node* node = root;
for(char c: word){
node = node->next[c-'a'];
if(!node)
return false;
}
return node->term;
}
/** Returns if there is any word in the trie that starts with the given prefix. */
bool startsWith(string prefix) {
Node* node = root;
for(char c: prefix){
node = node->next[c-'a'];
if(!node)
return false;
}
return true;
}
};
/**
* Your Trie object will be instantiated and called as such:
* Trie obj = new Trie();
* obj.insert(word);
* bool param_2 = obj.search(word);
* bool param_3 = obj.startsWith(prefix);
*/
209. Minimum Size Subarray Sum
Array
- 双指针
class Solution {
public:
int minSubArrayLen(int s, vector<int>& nums) {
int len = nums.size()+1;
int sum = 0;
int j = 0;
for(int i=0; i<nums.size(); i++){
sum += nums[i];
while(j<i && sum-nums[j]>=s){
sum -= nums[j];
j++;
}
if(sum>=s)
len = min(len, i-j+1);
printf("%d->%d: %d %d\n",j,i,sum,len);
}
return len==nums.size()+1 ? 0 : len;
}
};
210. Course Schedule II
- 有其他实现方式
class Solution {
public:
vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {
vector<int> ans;
unordered_map<int,int> h;
function<bool(int)> f = [&](int x){
if(h[x]==1)
return true;
if(h[x]==2)
return false;
h[x] = 1;
for(auto p: prerequisites)
if(p.first==x){
if(f(p.second))
return true;
}
h[x] = 2;
ans.push_back(x);
return false;
};
for(int i=0; i<numCourses; i++)
if(f(i))
return {};
return ans;
}
};
211. Add and Search Word – Data structure design
- Trie (Prefix Tree)
- DFS
class WordDictionary {
struct TrieNode{
TrieNode* next[26];
bool term;
TrieNode(){
memset(this,0,sizeof(TrieNode));
}
};
TrieNode* root;
public:
/** Initialize your data structure here. */
WordDictionary() {
root = new TrieNode();
}
/** Adds a word into the data structure. */
void addWord(string word) {
auto node = root;
for(char c: word){
auto& next = node->next[c-'a'];
if(!next)
next = new TrieNode();
node = next;
}
node->term = true;
}
/** Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. */
bool search(string word) {
return _search(word.c_str(), root);
}
bool _search(const char* s, TrieNode* root){
for(;;)
if(*s==0){
return root->term;
}else if(*s=='.'){
for(int i=0; i<26; i++){
if(root->next[i] && _search(s+1, root->next[i]))
return true;
}
return false;
}else{
root = root->next[*s-'a'];
s++;
if(root==NULL)
return false;
}
}
};
/**
* Your WordDictionary object will be instantiated and called as such:
* WordDictionary obj = new WordDictionary();
* obj.addWord(word);
* bool param_2 = obj.search(word);
*/
212. Word Search II
class Solution {
struct TrieNode{
TrieNode* next[26];
const char* term;
TrieNode(){
memset(this,0,sizeof(TrieNode));
}
};
TrieNode* root = new TrieNode();
void insert(string& word) {
auto node = root;
for(char c: word){
auto& next = node->next[c-'a'];
if(!next)
next = new TrieNode();
node = next;
}
node->term = word.c_str();
printf("> %s\n",node->term);
}
public:
vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
for(auto& word: words){
insert(word);
}
int m = board.size();
int n = board[0].size();
vector<string> ans;
function<void(int,int,TrieNode*)> f = [&](int y, int x, TrieNode* node){
char c = board[y][x];
if(c==0)
return;
auto next = node->next[c-'a'];
if(next==NULL)
return;
if(next->term){
ans.push_back(next->term);
next->term = NULL;
}
board[y][x] = 0;
if(x>0)
f(y, x-1, next);
if(x+1<n)
f(y, x+1, next);
if(y>0)
f(y-1, x, next);
if(y+1<m)
f(y+1, x, next);
board[y][x] = c;
};
for(int y=0;y<m;y++){
for(int x=0;x<n;x++){
f(y, x, root);
}
}
return ans;
}
};
213. House Robber II
DP
- 参考
198. House Robber - 区别：环，分两种情况断开
- http://m.blog.csdn.net/article/details?id=50386750
class Solution {
int rob1(vector<int>& nums, int s, int n) {
if(n==1)
return nums[s];
vector<int> d(n);
d[0] = nums[s];
d[1] = max(d[0], nums[s+1]);
for(int i=2; i<n; i++)
d[i] = max(d[i-1], d[i-2]+nums[s+i]);
return d[n-1];
}
public:
int rob(vector<int>& nums) {
int n = nums.size();
if(n==0)
return 0;
if(n==1)
return nums[0];
return max(rob1(nums,0,n-1),rob1(nums,1,n-1));
}
};
class Solution {
int rob1(vector<int>& nums, int s, int n) {
if(n==1)
return nums[s];
int a = 0, b = 0;
for(int i=0; i<n; i++){
int c = max(b, a+nums[s+i]);
a = b;
b = c;
}
return b;
}
public:
int rob(vector<int>& nums) {
int n = nums.size();
if(n==0)
return 0;
if(n==1)
return nums[0];
return max(rob1(nums,0,n-1),rob1(nums,1,n-1));
}
};
214. Shortest Palindrome
- 题目要求在前面添加字符
- 前缀，KMP
- 动态规可以么？
- http://www.jianshu.com/p/787b0499d871
class Solution {
public:
string shortestPalindrome(string s) {
auto s2 = s;
reverse(s2.begin(),s2.end());
string m = s + "#" + s2;
int n = m.size();
int b[n+1];
int i = 0, j = -1;
b[i] = j;
while(i<n){
while(j>=0 && m[i]!=m[j])
j = b[j];
i++, j++;
b[i] = j;
}
return s2.substr(0, s.size()-b[n]) + s;
}
};
215. Kth Largest Element in an Array
Sort
- 题目是查找，用排序
- quicksort
- heap
class Solution {
int partition(vector<int>& A, int a, int b){
int pivot = A[b];
int i = a;
for(int j=a;j<b;j++)
if(A[j]<pivot){
swap(A[i], A[j]);
i++;
}
swap(A[i], A[b]);
return i;
}
int search(vector<int>& nums, int a, int b, int k){
if(a<=b){
int p = partition(nums, a, b);
int rest = b - p;
if(k-1==rest)
return nums[p];
else if(k<=rest)
return search(nums, p+1, b, k);
else
return search(nums, a, p-1, k-rest-1);
}
return -1;
}
public:
int findKthLargest(vector<int>& nums, int k) {
return search(nums, 0, nums.size()-1, k);
}
};
class Solution {
int partition(vector<int>& A, int a, int b){
int pivot = A[b];
int i = a;
for(int j=a;j<b;j++)
if(A[j]<pivot){
swap(A[i], A[j]);
i++;
}
swap(A[i], A[b]);
return i;
}
int search(vector<int>& nums, int a, int b, int k){
while(a<=b){
int p = partition(nums, a, b);
int rest = b - p;
if(k-1==rest){
return nums[p];
}else if(k<=rest){
a = p+1;
}else{
b = p-1;
k = k-rest-1;
}
}
return -1;
}
public:
int findKthLargest(vector<int>& nums, int k) {
return search(nums, 0, nums.size()-1, k);
}
};
216. Combination Sum III
- 拿前面有道题题小改一下就行
class Solution {
public:
vector<vector<int>> combinationSum3(int k, int n) {
vector<vector<int>> result;
int sum = 0;
vector<int> item;
function<void(int)> loop = [&](int i){
if(sum==n && item.size()==k){
result.push_back(item);
return;
}
if(sum>n || item.size()==k)
return;
for(int j=i;j<=9;j++){
sum += j;
item.push_back(j);
loop(j+1);
sum -= j;
item.pop_back();
}
};
loop(1);
return result;
}
};
217. Contains Duplicate
- 这题没意思，是前面的题目的简化版本
class Solution {
public:
bool containsDuplicate(vector<int>& nums) {
unordered_map<int,bool> h;
for(int x: nums){
if(h[x])
return true;
h[x] = true;
}
return false;
}
};
218. The Skyline Problem
- 排序拐点
- 横坐标相同的点先插后删除
class Solution {
public:
vector<pair<int, int>> getSkyline(vector<vector<int>>& buildings) {
vector<pair<int, int>> ans;
vector<pair<int, int>> w;
for(auto& x: buildings){
w.push_back({x[0],-x[2]});
w.push_back({x[1],x[2]});
}
sort(w.begin(),w.end());
multiset<int> tree = {0};
int prev = 0;
for(auto& x: w){
if(x.second<0)
tree.insert(-x.second);
else
tree.erase(tree.find(x.second));
int top = *tree.rbegin();
if(top!=prev){
ans.push_back({x.first, top});
prev = top;
};
}
return ans;
}
};
219. Contains Duplicate II
class Solution {
public:
bool containsNearbyDuplicate(vector<int>& nums, int k) {
unordered_map<int,int> h;
for(int i=0; i<nums.size(); i++){
auto it = h.find(nums[i]);
if(it!=h.end() && i-it->second<=k)
return true;
h[nums[i]] = i;
}
return false;
}
};
220. Contains Duplicate III
- 滑动窗口
- 用搜索树，或者哈希表
class Solution {
public:
bool containsNearbyAlmostDuplicate(vector<int>& nums, int k, int t) {
multiset<long long> tree;
for(int i=0; i<nums.size(); i++){
long x = nums[i];
if(i-k-1>=0)
tree.erase(tree.lower_bound(nums[i-k-1]));
auto it = tree.lower_bound(x-t);
if(it!=tree.end() && distance(it, tree.upper_bound(x+t))>0)
return true;
tree.insert(nums[i]);
}
return false;
}
};
- lower_bound 返回大于等于的元素
class Solution {
public:
bool containsNearbyAlmostDuplicate(vector<int>& nums, int k, int t) {
multiset<long long> tree;
for(int i=0; i<nums.size(); i++){
long x = nums[i];
if(i-k-1>=0)
tree.erase(tree.lower_bound(nums[i-k-1]));
auto it = tree.lower_bound(x-t);
if(it!=tree.end() && *it<=x+t)
return true;
tree.insert(nums[i]);
}
return false;
}
};
221. Maximal Square
- 挺标准的动态规划的题目
- 非常标准
- 画图推演一下
class Solution {
public:
int maximalSquare(vector<vector<char>>& matrix) {
int m = matrix.size();
if(m==0)
return 0;
int n = matrix[0].size();
int ans = 0;
vector<int> d(n, 0);
for(int x=0;x<n;x++)
if(matrix[0][x]-'0')
ans = d[x] = 1;
for(int y=1;y<m;y++){
int c = d[0];
d[0] = matrix[y][0]-'0';
ans = max(ans, d[0]);
for(int x=1;x<n;x++){
int t = d[x];
if(matrix[y][x]-'0'){
d[x] = min(c, min(d[x], d[x-1]))+1;
ans = max(ans, d[x]);
}else{
d[x] = 0;
}
c = t;
}
}
return ans*ans;
}
};
222. Count Complete Tree Nodes
- 很容易超时
- 利用题目中完全二叉树的性质做优化
- 满二叉树有 2^k-1 个节点
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int countNodes(TreeNode* root) {
int d1 = 0, d2 = 0;
for(auto p=root; p; p=p->left)
d1++;
for(auto p=root; p; p=p->right)
d2++;
if(d1==d2)
return pow(2, d1)-1;
else
return countNodes(root->left)+countNodes(root->right)+1;
}
};
223. Rectangle Area
- 计算两个矩形的总面积
- 可能有重叠部分
class Solution {
public:
int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
int I = max(A,E), J = max(B,F), K= min(C,G), L=min(D,H);
return (C-A)*(D-B)+(G-E)*(H-F)-( K>I && L>J ?(K-I)*(L-J) : 0);
}
};
224. Basic Calculator
- 题目给的是比较特殊的情况
- 教科书 Stack/Queue 的章节有个中缀转后缀表达式的例子
- 暂时不需要用编译原理中写 Parser 的算法
- 题目假设输入都是有效的，省略了错误情况的处理
- 分母（是分母！）为 0，括号不匹配，token 重复
- 参考资料
class Solution {
int priority(char c){
switch(c){
case '+':
case '-':
return 1;
case '*':
case '/':
return 2;
default:
return 0;
}
}
void operate(){
int y = nums.top();
nums.pop();
int x = nums.top();
nums.pop();
switch(ops.top()){
case '+':
nums.push(x+y);
break;
case '-':
nums.push(x-y);
break;
case '*':
nums.push(x*y);
break;
case '/':
nums.push(x/y);
break;
}
ops.pop();
}
stack<int> nums;
stack<char> ops;
public:
int calculate(string s) {
char c;
const char *p = s.c_str();
while((c=*p)){
printf("> %c\n",c);
switch(c){
case ' ':
p++;
break;
case '+':
case '-':
case '*':
case '/':
while(!ops.empty() && priority(c)<=priority(ops.top()))
operate();
ops.push(c);
p++;
break;
case '(':
ops.push(c);
p++;
break;
case ')':
while(ops.top()!='(')
operate();
ops.pop();
p++;
break;
default:
char *e;
int x = strtol(p, &e, 10);
nums.push(x);
p = e;
}
}
while(!ops.empty())
operate();
return nums.top();
}
};
- 没有乘除法，只有加减和正整数，可以简化
- 求和，带符号
class Solution {
public:
int calculate(string s) {
int ans = 0;
stack<int> signs;
signs.push(1);
int sign = 1;
const char *p = s.c_str();
char c;
for(;;){
switch(*p){
case 0:
goto end_for;
case ' ':
p++;
break;
case '+':
sign = 1;
p++;
break;
case '-':
sign = -1;
p++;
break;
case '(':
signs.push(sign*signs.top());
sign = 1;
p++;
break;
case ')':
signs.pop();
sign = 1;
p++;
break;
default:
char *e;
int x = strtol(p, &e, 10);
ans += signs.top()*sign*x;
p = e;
}
}
end_for:
return ans;
}
};
225. Implement Stack using Queues
- 可以用 size(queue)
- 出入有一个操作为 O(N)
- 可以不需要额外空间
class MyStack {
queue<int> q1, q2;
public:
/** Initialize your data structure here. */
MyStack() {
}
/** Push element x onto stack. */
void push(int x) {
q2.push(x);
}
/** Removes the element on top of the stack and returns that element. */
int pop() {
while(q2.size()>1){
q1.push(q2.front());
q2.pop();
}
int y = q2.front();
q2.pop();
q1.swap(q2);
return y;
}
/** Get the top element. */
int top() {
int y = pop();
q2.push(y);
return y;
}
/** Returns whether the stack is empty. */
bool empty() {
return q2.empty();
}
};
/**
* Your MyStack object will be instantiated and called as such:
* MyStack obj = new MyStack();
* obj.push(x);
* int param_2 = obj.pop();
* int param_3 = obj.top();
* bool param_4 = obj.empty();
*/
226. Invert Binary Tree
- 二叉树遍历
class Solution {
public:
TreeNode* invertTree(TreeNode* root) {
if(root==nullptr)
return nullptr;
swap(root->left, root->right);
invertTree(root->left);
invertTree(root->right);
return root;
}
};
227. Basic Calculator II
- 前面那题的特例/子集
- 不叫优化，叫简化/特殊化
int calculate(char* s) {
int ans = 0;
int x = strtol(s,&s,10);
while(*s){
while(isspace(*s))
s++;
char op = *s++;
int y = strtol(s,&s,10);
switch(op){
case '+':
ans += x;
x = y;
break;
case '-':
ans += x;
x = -y;
break;
case '*':
x *= y;
break;
case '/':
x /= y;
break;
}
}
ans += x;
return ans;
}
class Solution {
public:
int calculate(string s) {
stack<int> st;
int x;
const char *p = s.c_str();
st.push(strtol(p,(char**)&p,10));
while(*p){
while(isspace(*p))
p++;
char op = *p++;
int x, y = strtol(p,(char**)&p,10);
switch(op){
case '+':
st.push(y);
break;
case '-':
st.push(-y);
break;
case '*':
x = st.top();
st.pop();
st.push(x*y);
break;
case '/':
x = st.top();
st.pop();
st.push(x/y);
break;
}
}
int ans = 0;
while(!st.empty()){
ans += st.top();
st.pop();
}
return ans;
}
};
228. Summary Ranges
class Solution {
public:
vector<string> summaryRanges(vector<int>& nums) {
vector<string> ans;
stringstream b;
int s = 0;
for(int i=1; s!=nums.size(); i++){
if(i==nums.size() || nums[i]!=nums[i-1]+1){
if(s==i-1){
ans.push_back(to_string(nums[i-1]));
}else{
stringstream b;
b << nums[s] << "->" << nums[i-1];
ans.push_back(b.str());
}
s = i;
}
}
return ans;
}
};
229. Majority Element II
class Solution {
public:
vector<int> majorityElement(vector<int>& nums) {
int a = 0, ca = 0;
int b = 0, cb = 0;
for(int x: nums){
if(x==a){
ca++;
}else if(x==b){
cb++;
}else if(ca==0){
a = x, ca = 1;
}else if(cb==0){
b = x, cb = 1;
}else{
cb--, ca--;
}
}
ca = cb = 0;
for(int x: nums){
if(x==a)
ca++;
else if(x==b)
cb++;
}
vector<int> ans;
if(ca>nums.size()/3)
ans.push_back(a);
if(cb>nums.size()/3)
ans.push_back(b);
return ans;
}
};
230. Kth Smallest Element in a BST
Tree
- 这题本身没什么，附加问挺有意思的
- 依然基于排序二叉树的模版，针对题目做适当的优化
- 在二叉树添加和查找的时候，在每个节点维持做节点个数。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
int count = 0;
int ans;
bool f(TreeNode* root){
if(!root)
return false;
if(f(root->left))
return true;
count--;
if(count==0){
ans = root->val;
return true;
}
if(f(root->right))
return true;
return false;
}
public:
int kthSmallest(TreeNode* root, int k) {
count = k;
f(root);
return ans;
}
};
231. Power of Two
- 是 2, 4, 8, 16, 32, 64 …
- 利用二进制中 1 的个数的那题，判断最高位
class Solution {
public:
bool isPowerOfTwo(int n) {
return n>0 && !(n&(n-1));
}
};
232. Implement Queue using Stacks
Stack
- 在函数式不可变的数据结构中用得到
class Queue {
stack<int> s1;
stack<int> s2;
void enq(){
while(!s2.empty()){
s1.push(s2.top());
s2.pop();
}
}
public:
// Push element x to the back of queue.
void push(int x) {
s2.push(x);
}
// Removes the element from in front of queue.
void pop(void) {
if(s1.empty())
enq();
s1.pop();
}
// Get the front element.
int peek(void) {
if(s1.empty())
enq();
return s1.top();
}
// Return whether the queue is empty.
bool empty(void) {
return s1.empty() && s2.empty();
}
};
233. Number of Digit One
Integer
class Solution {
public:
int countDigitOne(int n) {
int a=0, b=1, c=1;
while(n){
a += (n + 8) / 10 * b;
if(n%10==1)
a += c;
c += (n%10)*b;
b *= 10;
n /= 10;
}
return a;
}
};
234. Palindrome Linked List
- 拿前面有道链表题改改就行
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
ListNode* cut(ListNode* head){
if(!head)
return NULL;
auto slow = head, fast = head;
while(fast && fast->next && fast->next->next){
slow = slow->next;
fast = fast->next->next;
}
auto head2 = slow->next;
slow->next;
return head2;
}
ListNode* reverse(ListNode* head){
ListNode* prev = NULL;
while(head){
auto next = head->next;
head->next = prev;
prev = head;
head = next;
}
return prev;
}
public:
bool isPalindrome(ListNode* head) {
auto head2 = reverse(cut(head));
for(auto p=head,q=head2; p&&q; p=p->next,q=q->next)
if(p->val!=q->val)
return false;
return true;
}
};
235. Lowest Common Ancestor of a Binary Search Tree
- 已知根节点
- 利用二叉排序树的节点值的关系
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
for(;;)
if(root->val > p->val && root->val > q->val){
root = root->left;
}else if(root->val < p->val && root->val < q->val){
root = root->right;
}else{
return root;
}
}
};
236. Lowest Common Ancestor of a Binary Tree
- 递归，验证左右节点是否为父节点
- 都不是，则自己是
- 有一个是，在那边
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if(!root || root==p || root==q)
return root;
auto left = lowestCommonAncestor(root->left, p, q);
auto right = lowestCommonAncestor(root->right, p, q);
if(left && right)
return root;
return left ? left : right;
}
};
237. Delete Node in a Linked List
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
void deleteNode(ListNode* node) {
auto next = node->next;
assert(next);
*node = *next;
delete next;
}
};
238. Product of Array Except Self
- 题目要求不用除法，那就左右来回两趟
- 左边乘以右边
class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
int n = nums.size();
vector<int> ans(n);
int c = 1;
for(int i=0; i<n; i++){
ans[i] = c;
c *= nums[i];
}
c = 1;
for(int i=n-1; i>=0; i--){
ans[i] *= c;
c *= nums[i];
}
return ans;
}
};
239. Sliding Window Maximum
- 求滑动窗口中的最大值序列
- 用 deque 可以 实现 O(N)
- 用 heap O(N LOG K)，直接实现 O(KN)
- http://articles.leetcode.com/sliding-window-maximum/
- http://www.cnblogs.com/yrbbest/p/5004596.html
class Solution {
public:
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
if(nums.size()<k || k==0)
return {};
vector<int> m;
deque<int> q;
for(int i=0; i<k; i++){
while(!q.empty() && nums[i]>=nums[q.back()])
q.pop_back();
q.push_back(i);
}
for(int i=k; i<nums.size(); i++){
m.push_back(nums[q.front()]);
while(!q.empty() && nums[i]>=nums[q.back()])
q.pop_back();
while(!q.empty() && q.front()<=i-k)
q.pop_front();
q.push_back(i);
}
m.push_back(nums[q.front()]);
return m;
}
};
240. Search a 2D Matrix II
- 从右上角/左下角开始搜索，往左下/右上方向进行
class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int m = matrix.size();
if(m==0)
return false;
int n = matrix[0].size();
int x = n-1, y=0;
while(y<m && x>=0){
if(matrix[y][x]==target){
return true;
}if(matrix[y][x]<target)
y++;
else{
x--;
}
}
return false;
}
};
241. Different Ways to Add Parentheses
Search
- 表达式语法树，这里是二叉树
- 每个符号都有机会做根节点
- 不要遗漏解，挺容易遗漏的
- 求全部解，笛卡尔积
- 递归函数返回数组，或者使用回调
- 存在左右子树两个方向，都各有若干可能
- 这是这道题有趣的地方
class Solution {
const char* s;
void f(int a, int b, function<void(int)> c){
bool flag = false;
for(int i=a; i<=b; i++){
switch(s[i]){
case '+':
flag = true;
f(a,i-1,[&](int x){
f(i+1,b,[&](int y){
c(x+y);
});
});
break;
case '-':
flag = true;
f(a,i-1,[&](int x){
f(i+1,b,[&](int y){
c(x-y);
});
});
break;
case '*':
flag = true;
f(a,i-1,[&](int x){
f(i+1,b,[&](int y){
c(x*y);
});
});
break;
}
}
if(!flag){
c(strtol(s+a,0,10));
}
}
public:
vector<int> diffWaysToCompute(string input) {
vector<int> ans;
this->s = input.c_str();
f(0,input.size()-1,[&](int z){
ans.push_back(z);
});
return ans;
}
};
242. Valid Anagram
- 排序，或者哈希
class Solution {
public:
bool isAnagram(string s, string t) {
sort(s.begin(), s.end());
sort(t.begin(), t.end());
return s==t;
}
};
class Solution {
public:
bool isAnagram(string s, string t) {
unordered_map<char,int> h;
for(char x: s)
h[x]++;
for(char x: t)
h[x]--;
for(auto p: h)
if(p.second)
return false;
return true;
}
};
索引
归类
解答 1.5
有锁的题放这里
解答 2
以下用来尝试一些不那么好的解法。
有的是不符合题目的全部要求，有的就是随手写写的。
部分题目的解答
用 Ruby
扩展想法用，写一些并不一定完全复合题目要求。
注意
- Ruby 负数除法的取整方向和 C 不一样
1. Two Sum
- array + hash
def two_sum(nums, target)
h = {}
for x, i in nums.each_with_index
return h[x], i if h.key? x
h[target-x] = i
end
end
2. Add Two Numbers
def add_two_numbers(l1, l2)
list = node = ListNode.new(nil)
c = 0
while l1 || l2 || c!=0
if l1
c += l1.val
l1 = l1.next
end
if l2
c += l2.val
l2 = l2.next
end
node.next = ListNode.new(c%10)
c /= 10
node = node.next
end
list.next
end
3. Longest Substring Without Repeating Characters
# @param {String} s
# @return {Integer}
def length_of_longest_substring(s)
ans = 0
h = Hash.new(-1)
j = -1
for c, i in s.each_char.with_index
j = [j, h[c]].max
h[c] = i
ans = [ans, i-j].max
end
ans
end
17. Letter Combinations of a Phone Number
- 回溯
- 这题每一步不需要加条件，后面的回溯题会先过滤路径上的点。
- 回溯的每一项是“或”的关系
# @param {String} digits
# @return {String[]}
def letter_combinations(digits)
h = ["","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"]
f = ->(s){
if s.empty?
[""]
else
h[s[0].to_i].chars.flat_map{|x|
f.(s[1..-1]).map{|y|
x+y
}
}
end
}
digits.empty? ? [] : f.(digits)
end
# @param {String} digits
# @return {String[]}
def letter_combinations(digits)
h = ["","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"]
f = ->(i){
if i==digits.size
[""]
else
h[digits[i].to_i].chars.flat_map{|x|
f.(i+1).map{|y|
x+y
}
}
end
}
digits.empty? ? [] : f.(0)
end
上面这个以后删掉
# @param {String} digits
# @return {String[]}
def letter_combinations(digits)
return [] if digits.empty?
h = ["","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"]
ans = []
path = []
f = ->(i){
if i==digits.size
ans.push path*''
else
for c in h[digits[i].to_i].chars
path.push c
f.(i+1)
path.pop
end
end
}
f.(0)
ans
end
写成用 Stack 的形式
# @param {String} digits
# @return {String[]}
def letter_combinations(digits)
h = ["","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"]
ans = []
stack = []
stack.push(-1)
while !stack.empty?
stack[-1] += 1
p stack
if stack[-1]>=h[digits[stack.size-1].to_i].size
stack.pop
elsif stack.size==digits.size
ans.push stack.map.with_index{|x,i|
h[digits[i].to_i][x]
}*''
else
stack.push(-1)
end
end
ans
end
def letter_combinations(digits)
h = ["","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"]
return [] if digits.empty?
ans = []
stack = []
stack.push(-1)
path = []
while x = stack.pop
if stack.size==digits.size
ans.push path*''
path.pop
next
end
x += 1
if x>=h[digits[stack.size].to_i].size
path.pop
else
path.push(h[digits[path.size].to_i][x])
stack.push(x)
stack.push(-1)
end
end
ans
end
def letter_combinations(digits)
h = ["","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"]
return [] if digits.empty?
ans = []
stack = []
stack.push(-1)
path = []
while !stack.empty?
if stack.size>digits.size
ans.push path*''
path.pop
stack.pop
next
end
stack[-1] += 1
if stack[-1]>=h[digits[stack.size-1].to_i].size
path.pop
stack.pop
else
path.push(h[digits[path.size].to_i][stack[-1]])
stack.push(-1)
end
end
ans
end
def letter_combinations(digits)
h = ["","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"]
return [] if digits.empty?
ans = []
stack = []
stack.push(-1)
path = []
while !stack.empty?
if stack.size>digits.size
ans.push path*''
stack.pop
next
end
o = stack[-1]
stack[-1] += 1
if stack[-1]>=h[digits[stack.size-1].to_i].size
path.pop if o!=-1
stack.pop
else
path.pop if o!=-1
path.push(h[digits[path.size].to_i][stack[-1]])
stack.push(-1)
end
end
ans
end
38. Count and Say
def count_and_say(n)
(n-1).times.inject("1"){|a,b|a.chars.chunk(&:itself).map{|k,v|"#{v.size}#{k}"}*''}
end
49. Group Anagrams
def group_anagrams(strs)
strs.group_by{|x|x.chars.sort}.values
end
51. N-Queens
def solve_n_queens(n)
a = [nil]*n
b = [false]*n
c = [false]*(2*n-1)
d = [false]*(2*n-1)
ys = []
f = ->(i){
if i < n
for j in 0...n
unless b[j] || c[i+j] || d[i-j+n-1]
a[i] = j
b[j] = c[i+j] = d[i-j+n-1] = true
f.(i + 1)
b[j] = c[i+j] = d[i-j+n-1] = false
end
end
else
ys << a.collect{|x|
line = "."*n
line[x] = ?Q
line
}
end
}
f[0]
ys
end
65. Valid Number
def is_number(s)
s=~/^\s*[-+]?(\d+(\.\d*)?|\.\d+)([eE][-+]?\d+)?\s*$/ ? true : false;
end
66. Plus One
def plus_one(digits)
(digits.size-1).downto(0) do |i|
if(digits[i]!=9)
digits[i] += 1
return digits
end
digits[i] = 0
end
digits.unshift(1)
return digits
end
78. Subsets
def subsets(nums)
(1<<nums.length).times.map{|i|nums.select.with_index{|x,j|(i>>j)&1==1}}
end
94. Binary Tree Inorder Traversal
def inorder_traversal(root)
result = []
stack = []
node = root
while node || !stack.empty?
if node
stack << node
node = node.left
else
node = stack.pop
result << node.val
node = node.right
end
end
result
end
def inorder_traversal(root)
result = []
stack = []
node = root
while true
if node
stack << node
node = node.left
elsif !stack.empty?
node = stack.pop
result << node.val
node = node.right
else
break
end
end
result
end
102. Binary Tree Level Order Traversal
- 二叉树层次遍历
def level_order(root)
result = []
queue = []
queue << root if root
until queue.empty?
result << queue.map(&:val)
queue.size.times do
node = queue.shift
queue << node.left if node.left
queue << node.right if node.right
end
end
result
end
144. Binary Tree Preorder Traversal
def preorder_traversal(root)
result = []
stack = []
stack << root if root
while node = stack.pop
result << node.val
stack << node.right if node.right
stack << node.left if node.left
end
result
end
145. Binary Tree Postorder Traversal
def postorder_traversal(root)
result = []
stack = []
node = root
prev = nil
while true
if node
stack << node
node = node.left
elsif !stack.empty?
node = stack.last
if prev==node.right || node.right.nil?
result << node.val
prev, node = node, nil
stack.pop
else
node = node.right
end
else
break
end
end
result
end
208. Implement Trie (Prefix Tree)
class Trie
def initialize()
@root = {}
end
def insert(word)
node = @root
word.each_char do |c|
node = node[c]||={}
end
node[nil] = true
end
def search(word)
node = @root
word.each_char do |c|
node = node[c] or return false
end
node[nil]==true
end
def starts_with(prefix)
node = @root
prefix.each_char do |c|
node = node[c] or return false
end
true
end
end
217. Contains Duplicate
def contains_duplicate(nums)
h = {}
nums.each do |e|
return true if h[e]
h[e] = true
end
false
end
def contains_duplicate(nums)
nums.sort!
for i in 1...nums.size
return true if nums[i]==nums[i-1]
end
false
end
224. Basic Calculator
def calculate(s)
ans = 0
stack = [1]
sign = 1
s.scan(/\d+|[+\-()]/) do |x|
case x
when ?+
sign = stack.last
when ?-
sign = -stack.last
when ?(
stack << sign
sign = stack.last
when ?)
stack.pop
else
ans += sign*x.to_i
end
end
ans
end
227. Basic Calculator II
def calculate(s)
ans = 0
e = s.scan(/\d+|[+\-*\/]/)
x = e.shift.to_i
while op = e.shift
y = e.shift.to_i
case op
when ?+
ans += x
x = y
when ?-
ans += x
x = -y
when ?*
x *= y
when ?/
x = x.fdiv(y).to_i
end
end
ans += x
end
解答3
Java 和 JavaScript
分类
C++11
在 OJ 中 C++98 支持得最广泛
- 出于速度和输出格式的考虑，用 printf
- 目前 LeetCode 中用的是 C++11
LeetCode 的有些题目是针对 C++/Java 设计的
C 中字符串指针在 C++ 中只读时可以使用，可以转换为用下标
- 数组也可以用指针，vector 只能用下标了。
- C 的字符串后缀是递归的数据结构
比 C++98 多了 hashtable
- 有的情况可以用排序
- 二分法查找
- 重复元素相邻排列
- 或者用数组
- key 少的时候用下标
- 线性探测法，特别是不删，有限的时也挺简单
- 有的情况可以用排序
vector尽量预先分配大小
有 move 之后可以直接返回 vector 了
比 Java 多了引用，链表问题中 next 的指针可以转换为添加一个头节点
和 Python 相比，有大括号和分号
内存管理，C++ 没有 Java 的 GC，相比 C 有 C++ 可以利用作用域
- 算法题可以利用程序结束释放所有内存
IO 函数 C 比 C++ 快，所以为了充分利用运行时间，尽量用 C 的
死循环或者复杂度太高会超时，下标越界或空指针会运行时错误
- 越界错误有时候本地会继续运行会未报错
STL
- C 的 string，qsort，bsearch
假设 Accepted 的解答为结果正确的解答
- LeetCode 有的题目在提交后会又有改动
- 原先 C++ 参数传数组改为用 STL
- 有个解答原来不超时的变得超时了
freebsd 可以看一些标准库的实现
感觉最初 LeetCode 是一个可以刷题的 Blog
- 老文 URL 改了，还可以看，感觉质量比更高
- http://articles.leetcode.com/
提交结果
- Runtime Error 越界，爆栈，空指针
- 后来开始返回错误信息了。。。
- Time Limit Exceeded 复杂度，死循环
- Runtime Error 越界，爆栈，空指针
参考资料
- http://en.cppreference.com
- LeetCode题解（灵魂机器） http://github.com/soulmachine/leetcode
- 有的题目给出多种思路，并且把同类型的解答归类整理
- 通过基础问题的扩展和组合可以应对更多的新的问题
- MaskRay 的解答 http://github.com/MaskRay/LeetCode/
- 技巧性非常强解法
- LeetCode 的博客有一些解答，部分题目后面也有解答
- 老的首页上的文章，目前在 http://articles.leetcode.com/
- http://www.programcreek.com/2012/11/top-10-algorithms-for-coding-interview/
- http://www.geeksforgeeks.org/
- http://bookshadow.com/leetcode/
- Python 为主
- http://blog.csdn.net/qq508618087/article/category/5910619
- http://www.cnblogs.com/grandyang/p/4606334.html
删除 http://algorithm.yuanbin.me/zh-hans/ 这个不太好。
其他
一些解答
其他解答
没细看，或者不太好的解答
LeetCode 的变化
- 以前是不返回错误信息的。
注意
- 考虑边界情况和异常情况
- 递归溢出
- 自行设计测试用例，更多情况不像 LeetCode 有提供的可以依赖
- 死循环的时候 Debug
- 使用 printf scanf
- 分析题干中的要求，运用基本的模型，分解为步骤的组合
- 有时候题目对输入有限定
- 用符合题目要求的，比较好实现的解答就可以了
- 坑
- .size() 为 unsigned，-1 会不对
整数
- 32位补码有符号整数范围 $[-2{31},2{31}-1]$ ，最小负数取相反数会溢出
- 在 C 中负数取模为负数，不同语言有不同规则
- 负数右移是除以2向下取整数，除法是向零取整数。
- 有符号，无符号位移
数组/链表
数组和链表都可以按顺序访问
- 数组还可以按下标随机读写
- 链表有节点指针变换的问题
插入排序，选择排序
链表
- 有时候可以构建一个额外的头结点
- 注意元素个数为 0, 1, 2 的情况
栈，队列
- 可以用数组和链表实现
二叉树
- 递归遍历能应对大部分问题
图
有些状态机可以作为图
深度优先搜索用 Stack，广度优先搜索用 Queue
- 取出元素时处理
- 将相邻节点插入
- 有环图已搜索过的相邻节点不再插入
最短路径
- 举例为 1 时 变成广度优先搜索
- 使用优先级队列
查找
- 二分查找
排序
递归/回溯
- 典型问题
- 八皇后
- 排列组合
- 深度优先搜索
- 短路可以通过检查返回值快速返回
- 递归用栈写成迭代形式，并不能降低复杂度。
- 对提交代码的结果没有影响，除非题目对实现方案有特定要求
逻辑/搜索
排列组合
- 使用回溯法解答，通常写成递归形式
- 集合是无序的
- 全排列中升序的序列构成组合
- 有重复元素时，重复元素的顺序改变同一个排列
- 集合元素可以用转换为用下标（索引）表示
动态规划
重叠
字符串
- 作为数组问题处理
- 实现 strstr 函数
质数
- 因式分解
C 语言
有些 C 语言的问题没在数据结构书中出现，也有出现的
- memmove
