LeetCode之Palindrome Partitioning_hihocoder x is a palindrome |x|<=|str| we can find-优快云博客

本文链接：https://blog.youkuaiyun.com/shiyimin1/article/details/48766881

本文介绍了一种使用深度优先搜索(DFS)实现的字符串回文划分算法。通过两种不同的递归方法展示了如何将一个字符串分割成多个子串，并确保每个子串都是回文串。文章详细解释了递归调用的过程及回文判断条件。

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/*采用深度搜索算法。
写法一。*/
class Solution {
public:
	vector<vector<string>> partition(string s) {
		vector<vector<string> > res;
		if(s.empty()) return res;
		vector<string> path;
		partition(s, res, path, -1, 0);
		return res;
	}

	void partition(const string &s, vector<vector<string> > &res, vector<string> &path, int pre, int cur){
		if(cur == s.size()-1){
			if(is_palindrome(s, pre+1, cur)){
				path.push_back(s.substr(pre+1, cur-pre));
				res.push_back(path);
				path.pop_back();
			}
			return;
		}
		else{
			if(is_palindrome(s, pre+1, cur)){
				path.push_back(s.substr(pre+1, cur-pre));
				partition(s, res, path, cur, cur+1);
				path.pop_back();
			}
			partition(s, res, path, pre, cur+1);
		}
	}

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

/*写法二。*/
class Solution {
public:
	vector<vector<string>> partition(string s) {
		vector<vector<string> > res;
		if(s.empty()) return res;
		vector<string> path;
		partition(s, res, path, 0);
		return res;
	}

	void partition(const string &s, vector<vector<string> > &res, vector<string> &path, int cur){
		if(cur == s.size()){
			res.push_back(path);
			return;
		}
		else{
			for(int i = cur; i < s.size(); ++i){
				if(is_palindrome(s, cur, i)){
					path.push_back(s.substr(cur, i-cur+1));
					partition(s, res, path, i+1);
					path.pop_back();
				}
			}
		}
	}

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