LeetCode 131 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.

A palindrome string is a string that reads the same backward as forward.

Solution

将字符串分割为所有可能的回文串。 首先用 \(check\) 来判断是否为回文串。接着用回溯来进行 \(dfs\)：对当前位置 \(pos\) 和下一个位置 \(i\) 之间的字符串来判断是否为回文串，接着从 \(i+1\) 处开始

点击查看代码

class Solution {
private:
    vector<vector<string>> ans;
    vector<string> res;
    
    bool check(string x, int st, int ed){
        while(st<=ed){
            if(x[st++]!=x[ed--])return false;
        }
        return true;
    }
    
    void dfs(string s, int pos, vector<string>& res){
        if(pos==s.size()){
            ans.push_back(res);return;
        }
        for(int i=pos;i<s.size();i++){
            string tmp = s.substr(pos,i-pos+1);
            if(check(s, pos, i)){
                res.push_back(tmp);
                dfs(s, i+1, res);
                res.pop_back();
            }
        }
    }
    
public:
    vector<vector<string>> partition(string s) {
        dfs(s, 0, res);
        return ans;
    }
};