[LeetCode] Palindrome Partitioning-优快云博客

本文链接：https://blog.youkuaiyun.com/a83610312/article/details/12455499

本文介绍了一种使用回溯算法和动态规划解决字符串回文划分问题的方法。通过构建DP数组判断子串是否为回文，再利用回溯算法寻找所有可能的回文划分组合。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Palindrome Partitioning

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"]
  ]

我居然是先做了II题，呵呵，一样的，有了上一题的dp数组来帮助判断某一个子串是否是回文了之后，我们回溯一下构造解就好了。

class Solution {
public:
    vector<vector<string>> partition(string s) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int n=s.length();
		vector<vector<bool> > dp(n+1,vector<bool>(n,true));
		for(int len=2;len<=n;len++)
		{
			for(int start=0;start<=n-len;start++)
			{
				dp[len][start]=s[start]==s[len+start-1]&&dp[len-2][start+1];
			}
		}
		vector<vector<string> > ret;
		vector<string> part;
		solve(0,s,part,ret,dp);
		return ret;
    }
	void solve(int k,string& s,vector<string>& part,vector<vector<string> > & ret,vector<vector<bool> >& dp)
	{
		if(k>=s.length())
		{
			ret.push_back(part);
			return;
		}
		string cur;
		for(int i=k;i<s.length();i++)
		{
			cur.push_back(s[i]);
			if(dp[i-k+1][k])
			{
				part.push_back(cur);
				solve(i+1,s,part,ret,dp);
				part.pop_back();
			}
		}
	}
};