131. Palindrome Partitioning

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131. Palindrome Partitioning

题目

 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"]
]

解析

• The Idea is simple: loop through the string, check if substr(0, i) is palindrome. If it is, recursively call dfs() on the rest of sub string: substr(i+1, length). keep the current palindrome partition so far in the ‘path’ argument of dfs(). When reaching the end of string, add current partition in the result.

• 理解dfs的递归思想，经典！！

• 控制输出的顺序：牛客网的oj系统不合理，但是也值得思考！简单的reverse(ret.begin(), ret.end()); 逆序输出不对的。

对应输出应该为:

[["d","d","e"],["dd","e"]]

你的输出为:

[["dd","e"],["d","d","e"]]

```C++
// Palindrome(回文) Partitioning
class Solution_131_ref {
	// date 2017/12/29 11:14

	// 如果要求输出所有可能的解，往往都是要用深度优先搜索。如果是要求找出最优的解，或者解的数量，往往可以使用动态规划
	// Reference：https://leetcode.com/problems/palindrome-partitioning/discuss/41964

public:
	vector<vector<string>> partition(string s) {
		vector<vector<string> > ret;
		if (s.empty()) return ret;

		vector<string> path;
		dfs(0, s, path, ret);

		return ret;
	}

	void dfs(int index, string& s, vector<string>& path, vector<vector<string> >& ret) {
		if (index == s.size()) {
			ret.push_back(path);
			return;
		}
		for (int i = index; i < s.size(); ++i) { //先以步长为1找回文串，找到了下一个回文串也是以步长为1开始 ；下一轮循环以步长为2开始，这样保证所有子回文串找到
			if (isPalindrome(s, index, i)) {
				path.push_back(s.substr(index, i - index + 1));
				dfs(i + 1, s, path, ret);
				path.pop_back();
			}
		}
	}

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

};

class Solution_131{

public:

	bool ispalindrome(string str)
	{
		if (str.size()==1)
		{
			return true;
		}

		bool flag = true;
		for (int i = 0; i < str.size()/2; i++)
		{
			if (str[i]==str[str.size()-1-i])
			{
				continue;
			}
			else
			{
				flag = false;
			}
		}
		return flag;
	}

	bool isPalindrome1(string s){
		return s == string(s.rbegin(), s.rend());
	}
	void dfs(string src, vector<string> &path, vector<vector<string>> &ret){
		if (src.size() <= 0 )
		{
			return; //递归退出条件
		}

		if (ispalindrome(src)) //最后部分的子串为回文串，结束此处递归 //
		{
			path.push_back(src);
			ret.push_back(path);
			path.pop_back();
		}

		string temp;
		for (int j = 1; j < src.size();j++)  //限制了长度2以上；
		{
			temp = src.substr(0, j);
		
			if (ispalindrome(temp))
			{
				path.push_back(temp);
				dfs(src.substr(j),path,ret);
				//ret.push_back(path);
				  
				path.pop_back();// 
			}

		}
		return;
	}

	
	void dfs1(string s, vector<string> &cur, vector<vector<string>> &res){
		if (s == ""){
			res.push_back(cur);
			return;
		}

		for (int i = 1; i <= s.length(); ++i) {
			string sub = s.substr(0, i);
			if (ispalindrome(sub)){
				cur.push_back(sub);
				dfs1(s.substr(i, s.length() - i), cur, res);
				cur.pop_back();
			}
		}

	}

	vector<vector<string>> partition(string s) {

		vector<vector<string>> ret;
		vector<string > path;

		dfs1(s, path, ret); //深度递归

		return ret; 
	}
};

题目来源

• 131. Palindrome Partitioning
• LeetCode39:Combination Sum