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leetcode dfs Palindrome Partitioning

Palindrome Partitioning

 Total Accepted: 21056 Total Submissions: 81036My Submissions

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"]
  ]


题意:分割字符串,使每个子串都是回文
思路:dfs
选择一个分割点时,如果从起点到分割点的子串是回文,那么该分割点是合法的,可以选择
按这个规则dfs枚举就可以了
复杂度:时间O(n),空间O(log n)


vector<vector<string> >res;


bool is_palindrome(const string &s, int start, int end){
	while(start < end){
		if(s[start] != s[end - 1]) return false;
		++start; --end;
	}
	return true;
}


void dfs(const string &s, int cur, vector<string>& partitions){
	int size = s.size();
	if(cur == size){
		res.push_back(partitions);
	}
	for(int end = cur + 1; end <= size; ++end){
		if(is_palindrome(s, cur, end)){
			partitions.push_back(s.substr(cur, end - cur));
			dfs(s, end, partitions);
			partitions.pop_back();
		}
	}
}


vector<vector<string>> partition(string s) {
	vector<string> tem;
	dfs(s, 0, tem);
	return res;
}


leetcode dfs Palindrome Partitioning