[Leetcode]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"]
  ]

典型的Backtrcking（回溯法）。从头开始判断长为i的子串是否回文，如果是继续判断剩余部分。

上代码：

class Solution {
public:
	vector<vector<string>> result;
	vector<vector<string>> partition(string s) {
		vector<string> oneAns;
		palindrome(s, oneAns);
		return result;
	}
	void palindrome(string s,vector<string> oneAns){
		if (0 == s.size()) return;
		for (int i = 1; i <= s.size(); i++){
			string tmp = string(s.begin(), s.begin() + i);
			if (isPalindrome(tmp)){
				oneAns.push_back(tmp);
				if (i == s.size()){
					result.push_back(oneAns);
					return;
				}
				else{
					palindrome(string(s.begin() + i, s.end()), oneAns);
				}
				oneAns.pop_back();
			}
			
		}
	}
	bool isPalindrome(string s){
		int i = 0,j = s.size() - 1;
		while (i < j){
			if (s[i] != s[j]){
				return false;
			}
			i++;
			j--;
		}
		return true;
	}
};