131 Palindrome Partitioning [Leetcode]

題目內容：

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"]
  ]

題目分析：
使用回溯的方法做，也可以稱爲DFS的方法。先窮舉所有迴文分支下的每個分支所有的狀況，再添加到結果中去，再回溯回上一個節點。
代碼實現，時間8ms：

class Solution {
public:
    vector<vector<string>> partition(string s) {
        vector<vector<string>> result;
        vector<string> temp;
        backTracking(s, 0, result, temp);
        return result;
    }

    void backTracking(string &s, int index, vector<vector<string>> &result, vector<string> &temp) {
        if(index >= s.size()) {
            result.push_back(temp);
            return;
        }

        for(int i = index; i < s.size(); ++i) {
            if(isPalindrome(index, i, s)) {
                temp.push_back(s.substr(index, i-index+1));
                backTracking(s, i+1, result, temp);
                temp.pop_back();//backtrack
            }
        }
    }

    bool isPalindrome(int start, int end, string &s) {
        while(start < end) {
            if(s[start] != s[end])
                return false;
            ++start;
            --end;
        }
        return true;
    }
};