Palindrome Partitioning

 1 class Solution {
 2 public:
 3     vector<vector<string>> partition(string s) {
 4         vector<vector<string>> res;
 5         if(s.length() == 0) return res;
 6         vector<string> path;
 7         dfs(res,path,s,0);
 8     }
 9     void dfs(vector<vector<string>> & res, vector<string> & path, string s, int start){
10         if(start == s.length()) res.push_back(path);
11         for(int i = start; i < s.length(); i++){
12             if(isPalindrome(s,start,i)){
13                 path.push_back(s.substr(start,i-start+1));
14                 dfs(res,path,s,i+1);
15                 path.pop_back();
16             }
17         }
18     }
19     bool isPalindrome(string s, int start, int end){
20         if(start == end) return true;
21         int i = start, j = end;
22         for(; i <= (start + end)/2 && s[i] == s[j]; i++, j--);
23         if(i > (start+end)/2) return true;
24         return false;
25     }
26 };