Leetcode 131:Palindrome Partitioning

Given a strings, partitionssuch that every substring of the partition is a palindrome.

Return all possible palindrome partitioning ofs.

For example, givens="aab",
Return

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

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//利用深度優先搜索DFS，將字符串劃分為一組回文字符串
class Solution {
public:
	vector> partition(string s) {
		vector> result;
		vector path;
		dfs(s, path, result, 0, 1);
		return result;
	}

	//s[0,prev-1]之間已經處理，保證是回文串
	//prev表示s[prev-1]與s[prev]之間的空隙位置，start同理
	void dfs(string &s, vector& path, vector>&result, int prev, int start)
	{
		if (start == s.size())
		{ //最後一個隔板
			if (ispalindrome(s, prev, start - 1))
			{   //必須使用
				path.push_back(s.substr(prev, start - prev));
				result.push_back(path);
				path.pop_back();
			}
			return;
		}
		
		//不斷開
		dfs(s, path, result, prev, start + 1);
		//如果[prev,start-1]是回文，則可以斷開，也可以不斷開(上一行已經做了)
		if (ispalindrome(s, prev, start - 1))
		{
			//斷開
			path.push_back(s.substr(prev, start - prev));
			dfs(s, path, result, start, start + 1);
			path.pop_back();
		}
	}

	//判斷子串是否為回文
	bool ispalindrome(const string &s, int start, int end)
	{
		while (start < end && s[start] == s[end])
		{
			++start;
			--end;
		}
		return start >= end;
	}
};