LeetCode--palindrome-partitioning

最新推荐文章于 2025-03-16 18:59:45 发布

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本文介绍了一种使用深度优先搜索解决回文字符串分割问题的方法。通过递归算法，找到所有可能的分割方式使得每个子串都是回文串。示例代码清晰地展示了如何实现这一算法。

题目描述

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"]
  ]

分析：这种求子序列的分割问题，可以想到用深度优先搜索方法

class Solution {
public:   
    void dfs(string s,vector<string>&path,vector<vector<string>>&res){
         
        if(s.size() < 1){
            res.push_back(path);
            return;
        }
        for(int i = 0;i<s.size();i++){
            int begin = 0;
            int end = i;
            while(begin < end){
                if(s[begin] == s[end]){
                    begin++;
                    end--;
                }
                else
                    break;               
            }
            if(begin >= end){
                path.push_back(s.substr(0,i+1));
                dfs(s.substr(i+1), path,res);
                path.pop_back();
            }
        }          
    }
    vector<vector<string>> partition(string s) {
        vector<vector<string>>res;
        vector<string>path;
        dfs(s,path,res);
        return res;
    }
};