131. Palindrome Partitioning (Back-Track, DP)

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: 首先使用动态规划纪录从i到j是否是回文数；然后遍历字符串，对于dp[depth][i]有选择与不选择两种情况，所以使用带回溯的递归，回溯法注意在递归返回后要将递归前改动的内容复原。

class Solution {
public:
    void backTracking(string s, int depth, vector<vector<bool>>& dp, vector<string>& current){
         for(int i = depth; i <s.length(); i++){
             if(dp[depth][i]){
                 current.push_back(s.substr(depth, i-depth+1));
                 if(i==s.length()-1) ret.push_back(current);
                 else backTracking(s,i+1, dp,current);
                 current.pop_back(); //back track
             } 
         }
    }
    vector<vector<string>> partition(string s) {
        //dp[i][j]: s[i...j] is parlindrome
        //dp[i][j] = dp[i-1][j+1] && s[i]==s[j]
        //traverse order: shorter one should be checked first, like insert sort
        int len = s.length();
        vector<vector<bool>> dp(len, vector<bool>(len, false));
        vector<string> current;
        for(int i = 0; i < len; i++) dp[i][i]=true;
        for(int i = 1; i < len; i++){
            for(int j = 0; j < i; j++){ //traverse the length
                if(s[i]==s[j]){
                    if(j==i-1) dp[j][i] = true;
                    else dp[j][i]=dp[j+1][i-1];
                }
            }
        }
        backTracking(s, 0, dp, current);
        return ret;
    }
private: 
    vector<vector<string>> ret;
};