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"]
]

class Solution {
public:
    void DFS(string &s,vector<string> &path,vector<vector<string>> &result,int start)
    {
        if(start == s.size())
        {
            result.push_back(path);
            return;
        }
        for(int i=start;i<s.size();i++)
        {
            if(isPalindrome(s,start,i))
            {
                path.push_back(s.substr(start,i-start+1));
                DFS(s,path,result,i+1);//不是所有的字符都參與循環(huán)
                path.pop_back();
            }
        }
    }
    bool isPalindrome(string &s,int start,int end)
    {
        while(start < end&&s[start] == s[end])
        {
            start++;
            end--;
        }
        if(start>=end)
           return true;
        else
           return false;
    }

    vector<vector<string>> partition(string s) {
        vector<vector<string>> result;
        vector<string> path;
        DFS(s,path,result,0);
        return result;
    }
};