Leetcode 131:Palindrome Partitioning

Posted 2020-06-15 zhihua_bupt

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"]
  ]

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//利用深度优先搜索DFS，将字符串划分为一组回文字符串
class Solution {
public:
	vector<vector<string>> partition(string s) {
		vector<vector<string>> result;
		vector<string> path;
		dfs(s, path, result, 0, 1);
		return result;
	}

	//s[0,prev-1]之间已经处理，保证是回文串
	//prev表示s[prev-1]与s[prev]之间的空隙位置，start同理
	void dfs(string &s, vector<string>& path, vector<vector<string>>&result, int prev, int start)
	{
		if (start == s.size())
		{ //最后一个隔板
			if (ispalindrome(s, prev, start - 1))
			{   //必须使用
				path.push_back(s.substr(prev, start - prev));
				result.push_back(path);
				path.pop_back();
			}
			return;
		}
		
		//不断开
		dfs(s, path, result, prev, start + 1);
		//如果[prev,start-1]是回文，则可以断开，也可以不断开(上一行已经做了)
		if (ispalindrome(s, prev, start - 1))
		{
			//断开
			path.push_back(s.substr(prev, start - prev));
			dfs(s, path, result, start, start + 1);
			path.pop_back();
		}
	}

	//判断子串是否为回文
	bool ispalindrome(const string &s, int start, int end)
	{
		while (start < end && s[start] == s[end])
		{
			++start;
			--end;
		}
		return start >= end;
	}
};