题目：
Given a string s, partition s such that every substring of the partition is a palindrome.
Return all possible palindrome partitioning of s.
Example:
Input: "aab" Output: [ ["aa","b"], ["a","a","b"] ]代码：
方法一——DFS：
class Solution { public: vector<vector<string>> partition(string s) { vector<vector<string>> res; vector<string> out; helper(s, 0, out, res); return res; } void helper(string s, int start, vector<string>& out, vector<vector<string>>& res) { if (start == s.size()) { res.push_back(out); return; } for (int i = start; i < s.size(); ++i) { if (!isPalindrome(s, start, i)) continue; out.push_back(s.substr(start, i - start + 1)); helper(s, i + 1, out, res); out.pop_back(); } } bool isPalindrome(string s, int start, int end) { while (start < end) { if (s[start] != s[end]) return false; ++start; --end; } return true; } };方法二——动态规划+递归：
class Solution { public: vector<vector<string>> partition(string s) { int n = s.size(); vector<vector<string>> res; vector<string> out; vector<vector<bool>> dp(n, vector<bool>(n)); for (int i = 0; i < n; ++i) { for (int j = 0; j <= i; ++j) { if (s[i] == s[j] && (i - j <= 2 || dp[j + 1][i - 1])) { dp[j][i] = true; } } } helper(s, 0, dp, out, res); return res; } void helper(string s, int start, vector<vector<bool>>& dp, vector<string>& out, vector<vector<string>>& res) { if (start == s.size()) { res.push_back(out); return; } for (int i = start; i < s.size(); ++i) { if (!dp[start][i]) continue; out.push_back(s.substr(start, i - start + 1)); helper(s, i + 1, dp, out, res); out.pop_back(); } } };想法：多思考
