leetcode dfs Palindrome Partitioning

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

题意：切割字符串，使每一个子串都是回文
思路：dfs
选择一个切割点时，假设从起点到切割点的子串是回文，那么该切割点是合法的。能够选择
按这个规则dfs枚举就能够了
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;
}