Palindrome Partitioning问题及解法

问题描述：

Given a string s, partition s such that every substring of the partition is a palindrome.

Return all possible palindrome partitioning of s.

示例, given s = "aab",
Return

[  ["aa","b"],  ["a","a","b"]]

问题分析：

此类问题属于回溯法范畴，对每次取出的子串判断是否是回文的，若全都是，则把这些字串放入到res中，否则不作处理。

过程详见代码：

class Solution {public:    vector<vector<string>> partition(string s) {vector<vector<string>>res;vector<string> re;bl(res,re,0,s,s.length());        return res;}void bl(vector<vector<string>>&res, vector<string>& re, int start, string s,int left){if (!left){res.push_back(re);return;}for (int i = 1; i <= left; i++){string t = s.substr(start, i);if (!palindrome(t)) continue;re.push_back(t);bl(res, re, start + i, s, left - i);re.pop_back();}}bool palindrome(string s){for (int i = 0; i < s.length() / 2; i++){if (s[i] != s[s.length() - 1 - i]) return false;}return true;}};