Palindrome Partitioning问题及解法

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;
	}
};