Palindrome Partitioning

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"]
  ]

思路：

深搜

代码：

 1     bool isPalindrome(string s){
 2         int l = s.length();
 3         if(l == 1)
 4             return true;
 5         int i = 0;
 6         while(i < l/2){
 7             if(s[i] != s[l-i-1])
 8                 return false;
 9             i++;
10         }
11         return true;
12     }
13     void search(vector<vector<string> > &result, vector<string> &tmp, string s, int index){
14         if(index == s.length()){ 
15             result.push_back(tmp);
16             return;
17         }
18         for(int i = index+1; i <= s.length(); i++){
19             string t = s.substr(index, i-index);
20             if(isPalindrome(t)){
21                 tmp.push_back(t);
22                 search(result, tmp, s, i);
23                 tmp.erase(tmp.end()-1);
24             }
25         }
26     }
27     vector<vector<string> > partition(string s) {
28         // IMPORTANT: Please reset any member data you declared, as
29         // the same Solution instance will be reused for each test case.
30         vector<vector<string> > result;
31         vector<string> tmp;
32         search(result, tmp, s, 0);
33         return result;
34     }