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131. Palindrome Partitioning
题目
Given a string s, partition s such that every substring of the partition is a palindrome.
Return all possible palindrome partitioning of s.
For example, given s = "aab",
Return
[
["aa","b"],
["a","a","b"]
]
解析
The Idea is simple: loop through the string, check if substr(0, i) is palindrome. If it is, recursively call dfs() on the rest of sub string: substr(i+1, length). keep the current palindrome partition so far in the ‘path’ argument of dfs(). When reaching the end of string, add current partition in the result.
理解dfs的递归思想,经典!!
控制输出的顺序:牛客网的oj系统不合理,但是也值得思考!简单的
reverse(ret.begin(), ret.end());
逆序输出不对的。
对应输出应该为:
[[“d”,”d”,”e”],[“dd”,”e”]]
你的输出为:
[[“dd”,”e”],[“d”,”d”,”e”]]
“`
// Palindrome(回文) Partitioning
class Solution_131_ref {
// date 2017/12/29 11:14
// 如果要求输出所有可能的解,往往都是要用深度优先搜索。如果是要求找出最优的解,或者解的数量,往往可以使用动态规划
// Reference:https://leetcode.com/problems/palindrome-partitioning/discuss/41964
public:
vector<vector<string>> partition(string s) {
vector<vector<string> > ret;
if (s.empty()) return ret;
vector<string> path;
dfs(0, s, path, ret);
return ret;
}
void dfs(int index, string& s, vector<string>& path, vector<vector<string> >& ret) {
if (index == s.size()) {
ret.push_back(path);
return;
}
for (int i = index; i < s.size(); ++i) { //先以步长为1找回文串,找到了下一个回文串也是以步长为1开始 ;下一轮循环以步长为2开始,这样保证所有子回文串找到
if (isPalindrome(s, index, i)) {
path.push_back(s.substr(index, i - index + 1));
dfs(i + 1, s, path, ret);
path.pop_back();
}
}
}
bool isPalindrome(const string& s, int start, int end) {
while (start <= end) {
if (s[start++] != s[end--])
return false;
}
return true;
}
};
class Solution_131{
public:
bool ispalindrome(string str)
{
if (str.size()==1)
{
return true;
}
bool flag = true;
for (int i = 0; i < str.size()/2; i++)
{
if (str[i]==str[str.size()-1-i])
{
continue;
}
else
{
flag = false;
}
}
return flag;
}
bool isPalindrome1(string s){
return s == string(s.rbegin(), s.rend());
}
void dfs(string src, vector<string> &path, vector<vector<string>> &ret){
if (src.size() <= 0 )
{
return; //递归退出条件
}
if (ispalindrome(src)) //最后部分的子串为回文串,结束此处递归 //
{
path.push_back(src);
ret.push_back(path);
path.pop_back();
}
string temp;
for (int j = 1; j < src.size();j++) //限制了长度2以上;
{
temp = src.substr(0, j);
if (ispalindrome(temp))
{
path.push_back(temp);
dfs(src.substr(j),path,ret);
//ret.push_back(path);
path.pop_back();//
}
}
return;
}
void dfs1(string s, vector<string> &cur, vector<vector<string>> &res){
if (s == ""){
res.push_back(cur);
return;
}
for (int i = 1; i <= s.length(); ++i) {
string sub = s.substr(0, i);
if (ispalindrome(sub)){
cur.push_back(sub);
dfs1(s.substr(i, s.length() - i), cur, res);
cur.pop_back();
}
}
}
vector<vector<string>> partition(string s) {
vector<vector<string>> ret;
vector<string > path;
dfs1(s, path, ret); //深度递归
return ret;
}
};