[LeetCode] Palindrome Partitioning, Solution

最新推荐文章于 2025-09-30 16:36:44 发布

原创最新推荐文章于 2025-09-30 16:36:44 发布 · 142 阅读

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本文介绍了一种使用深度优先搜索(DFS)解决回文字符串分割问题的方法，通过递归地检查子串是否为回文来生成所有可能的分割方案。

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"]
  ]

» Solve this problem

[Thoughts]
这种需要输出所有结果的基本上都是DFS的解法。实现如下。

[Code]

1:       vector<vector<string>> partition(string s) {  
2:            vector<vector<string>> result;  
3:            vector<string> output;  
4:            DFS(s, 0, output, result);  
5:            return result;  
6:       }  
7:       void DFS(string &s, int start, vector<string>& output, vector<vector<string>> &result)  
8:       {      
9:            if(start == s.size())  
10:            {  
11:                 result.push_back(output);  
12:                 return;  
13:            }  
14:            for(int i = start; i< s.size(); i++)  
15:            {    
16:                 if(isPalindrome(s, start, i))  
17:                 {  
18:                      output.push_back(s.substr(start, i-start+1));  
19:                      DFS(s, i+1, output, result);  
20:                      output.pop_back();  
21:                 }  
22:            }  
23:       }  
24:       bool isPalindrome(string &s, int start, int end)  
25:       {  
26:            while(start< end)  
27:            {  
28:                 if(s[start] != s[end])  
29:                 return false;  
30:                 start++; end--;  
31:            }  
32:            return true;  
33:       }