[LeetCode] Palindrome Partitioning

最新推荐文章于 2019-09-02 10:48:14 发布

1.02^365的成长裂变

最新推荐文章于 2019-09-02 10:48:14 发布

阅读量3.5k

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode1

本文链接：https://blog.youkuaiyun.com/raojunyang/article/details/77925030

LeetCode1 专栏收录该内容

120 篇文章

订阅专栏

本文介绍了一种字符串回文分区的算法实现，通过递归深度优先搜索的方式寻找所有可能的回文子串组合，并提供了C++代码示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

[Problem]

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"]
  ]

[Analysis]
最简单的方法，直接深搜。更好一点的方法，用mark[i][j]记录s[i:j]是不是回文，然后再深搜。

[Solution]

class Solution {
public:
    // is the string a palindrome
    bool isPalindrome(string s){
        // string with one letter
        if(s.length() <= 1){
            return true;
        }
   
        // check
        int i = 0, j = s.length() - 1;
        while(i <= j){
            if(s[i] != s[j]){
                return false;
            }
            i++;
            j--;
        }
        return true;
    }
   
    // partition
    vector<vector<string> > partition(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
       
        vector<vector<string> > res;
   
        // empty string
        if(s.length() == 0){
            return res;
        }
        // string with one letter
        else if(s.length() == 1){
            vector<string> tmp;
            tmp.push_back(s);
            res.push_back(tmp);
        }
        // DFS
        else{
            // the length of the first part could be range from 0 to s.length()
            for(int i = 1; i <= s.length(); ++i){
                // get the first part
                string head = s.substr(0, i);
   
                // the first part is not a palindrome, continue
                if(!isPalindrome(head))continue;
   
                if(i == s.length()){
                    vector<string> tmp;
                    tmp.push_back(head);
                    res.push_back(tmp);
                }
                // generate the remained parts
                else{
                    vector<vector<string> > r = partition(s.substr(i, s.length() - i));
                    for(int j = 0; j < r.size(); ++j){
                        r[j].insert(r[j].begin(), head);
                        res.push_back(r[j]);
                    }
                }
            }
        }
        return res;
    }
};