二分查找各种变形,(剑指Offer)面试题38:数字在排序数组中出现的次数

  • 统计一个数字在排序数组中出现的次数。

#include<iostream>
#include<math.h>

#include <vector>
#include <string>
#include <deque>
#include <stack>
#include <queue>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>

#include <algorithm>
#include <functional>
#include <numeric> //accmulate

#include <iterator> //ostream_iterator
#include <fstream>
#include <iomanip>  //setprecision() setw()

#include <memory> 
#include <cstring> //memset
using namespace std;

//#define cin infile //一定不能再oj系统中,有错,导致超时等!!!
//C++文件输入
ifstream infile("in.txt", ifstream::in);

#include <limits>
#define INT_MIN     (-2147483647 - 1) /* minimum (signed) int value */
#define INT_MAX       2147483647    /* maximum (signed) int value */

#include <array>
#include <bitset>

// 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) {}
};


//1. 对于序列化:使用前序遍历,递归的将二叉树的值转化为字符,并且在每次二叉树的结点
//不为空时,在转化val所得的字符之后添加一个' , '作为分割。对于空节点则以 '#' 代替。
//2. 对于反序列化:按照前序顺序,递归的使用字符串中的字符创建一个二叉树(特别注意:
//在递归时,递归函数的参数一定要是char ** ,这样才能保证每次递归后指向字符串的指针会
//随着递归的进行而移动
class Solution_Serialize{
public:
    void help_seri(TreeNode* root, string &res)
    {
        if (!root)
        {
            res += '#,';
            return;
        }
        //前序遍历
        //char r[10];
        //sprintf(r, "%d", root->val);
        string r = to_string(root->val);
        res += r;
        res += ',';
        help_seri(root->left, res);
        help_seri(root->right, res);

    }

    char* Serialize(TreeNode *root) {

        if (!root)
        {
            return nullptr;
        }

        string res;

        help_seri(root, res);

        return const_cast<char*>(res.c_str());
    }

    // 由于递归时,会不断的向后读取字符串,所以一定要用**str,以保证得到递归后指针str指向未被读取的字符
    TreeNode* help_deseri(char** str)
    {
        if (**str=='#')
        {
            (*str)++;
            return nullptr;
        }
        int num = 0;
        while (**str != '\0'&&**str != ',')
        {
            num = num * 10 + (**str-'0');
            ++(*str);
        }
        TreeNode* node = new TreeNode(num);
        if (**str=='\0')
        {
            return node;
        }
        else
        {
            ++(*str);
        }

        node->left = help_deseri(str);
        node->right = help_deseri(str);

        return node;
    }

    TreeNode* Deserialize(char *str) {

        if (str==NULL)
        {
            return nullptr;
        }

        TreeNode* ret = help_deseri(&str);

        return ret;
    }
};


class Solution_37{

public:
    // 方法一:借助stl库函数
    int GetNumberOfK(vector<int> data, int k)
    {
        return count(data.begin(), data.end(), k);
    }

    // 方法二:O(n)遍历

    // 方法三:二分查找
    int GetNumberOfK_2(vector<int> data, int k) {
        int lower = getLower(data, k);
        int upper = getUpper(data, k);
        return upper - lower + 1;
    }

    // 获取k第一次出现的下标
    int getLower(vector<int> data, int k){
        int start = 0, end = data.size() - 1;
        int mid = (start + end) / 2;
        while (start <= end){
            if (data[mid] < k){  
                start = mid + 1;
            }
            else{  // data[mid] >= k
                end = mid - 1;
            }
            mid = (start + end) / 2;
        }
        return start;
    }
    // 获取k最后一次出现的下标
    int getUpper(vector<int> data, int k){
        int start = 0, end = data.size() - 1;
        int mid = (start + end) / 2;
        while (start <= end){
            if (data[mid] <= k){ // data[mid] == k
                start = mid + 1;
            }
            else{
                end = mid - 1;
            }
            mid = (start + end) / 2;
        }
        return end;
    }

    int getFirstK(int* data, int k, int start, int end){
        while (start <= end){
            int mid = start + ((end - start) >> 1);
            if (data[mid] == k){
                if ((mid > 0 && data[mid - 1] != k) || mid == 0)
                    return mid;
                else
                    end = mid - 1;
            }
            else if (data[mid] > k)
                end = mid - 1;
            else
                start = mid + 1;
        }
        return -1;
    }

    int getLastK(int* data, int length, int k, int start, int end){
        while (start <= end){
            int mid = start + ((end - start) >> 1);
            if (data[mid] == k){
                if ((mid < length - 1 && data[mid + 1] != k) || mid == length - 1)
                    return mid;
                else
                    start = mid + 1;
            }
            else if (data[mid] < k)
                start = mid + 1;
            else
                end = mid - 1;
        }
        return -1;
    }

    // 标准的二分查找
    int binary_search(int*data, int n, int target)
    {
        int low = 0, high = n - 1;
        while (low<=high)  //必须有等号,避免边界查找不到
        {
            int mid = low + (high - low) / 2;
            if (data[mid]==target)
            {
                return mid;
            }
            else if (data[mid]>target)
            {
                high = mid - 1;
            }
            else
            {
                low = mid + 1;
            }
        }
        return -1;
    }

    int binary_search_test(int*data, int n, int target)
    {
        int low = 0, high = n - 1;
        int mid = low + (high - low) / 2;
        while (low <= high)  //必须有等号,避免边界查找不到
        {        
            if (data[mid] == target)
            {
                return mid;
            }
            else if (data[mid] > target)
            {
                high = mid - 1;
            }
            else
            {
                low = mid + 1;
            }

            mid = low + (high - low) / 2;
        }
        return -1;
    }

    // 用二分法寻找上界
    int binary_search_upperbound(int array[], int low, int high, int target)
    {
        //Array is empty or target is larger than any every element in array 
        if (low > high || target >= array[high]) //控制找不到的情况
            return -1;

        int mid = (low + high) / 2;
        while (high > low)
        {
            if (array[mid] > target)
                high = mid;
            else
                low = mid + 1;

            mid = (low + high) / 2;
        }

        return mid;
    }
    int binary_search_lowerbound(int array[], int low, int high, int target)
    {
        //Array is empty or target is less than any every element in array
        if (high < low || target <= array[low]) 
            return -1;

        int mid = (low + high + 1) / 2; //make mid lean to large side
        while (low < high)
        {
            if (array[mid] < target)
                low = mid;
            else
                high = mid - 1;

            mid = (low + high + 1) / 2;
        }

        return mid;
    }


    int search_rotation(vector<int>& nums, int target) { //旋转数组查找
        int l = 0, r = nums.size() - 1;
        while (l <= r) {
            int mid = (l + r) / 2;
            if (target == nums[mid])
                return mid;
            // there exists rotation; the middle element is in the left part of the array
            if (nums[mid] > nums[r]) {
                if (target < nums[mid] && target >= nums[l])
                    r = mid - 1;
                else
                    l = mid + 1;
            }
            // there exists rotation; the middle element is in the right part of the array
            else if (nums[mid] < nums[l]) {
                if (target > nums[mid] && target <= nums[r])
                    l = mid + 1;
                else
                    r = mid - 1;
            }
            // there is no rotation; just like normal binary search
            else {
                if (target < nums[mid])
                    r = mid - 1;
                else
                    l = mid + 1;
            }
        }
        return -1;
    }
};

int main()
{
    int array[] = { 2, 3, 7, 7, 7, 13, 17 };
    int target = 7;

    Solution_37 su_37;
    int ret1=su_37.GetNumberOfK_2(vector<int>(array,array+7), target); //3

    int ret2 = su_37.binary_search(array,7, target); //3

    int ret3 = su_37.binary_search_test(array, 7, target); //3

    int ret4 = su_37.getUpper(vector<int>(array, array + 7), target); //4

    int ret5 = su_37.binary_search_upperbound(array, 0, 6, target);//5

    int ret6 = su_37.getLower(vector<int>(array, array + 7), target);//2

    int ret7 = su_37.binary_search_lowerbound(array, 0, 6, target);//1

    vector<int> res(array, array + 7);
    int ret8=lower_bound(res.begin(), res.end(), target)-res.begin(); //2

    int ret9 = upper_bound(array, array + 7, target)-array; //5


    Solution_Serialize su_1;
    char str[] = { '3', ',', '9', ',', '20', ',', ' #', '#', '15', ',', '7', ',' };
    TreeNode* ret = su_1.Deserialize(str);
    
    return 0;
}
    原文作者:ranjiewen
    原文地址: https://www.cnblogs.com/ranjiewen/p/9031633.html
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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