Leetcode_131. Palindrome Partitioning_[DFS]

题目链接

Given a string s, partition s such that every substring of the partition is a palindrome.

Return all possible palindrome partitioning of s.

Example:

Input: "aab"
Output:
[
  ["aa","b"],
  ["a","a","b"]
]

题意简单明了。
解法：要找到所有的划分方法，应想到穷举。

class Solution {
public:
    bool isPalindrome(string s){
        for(size_t i=0; i<s.size()/2; i++)
            if(s[i] != s[s.size()-1-i])
                return false;
        return true;
    }

    vector<vector<string>> partition(string s) {
        vector<vector<string>> ret;
        vector<string> temp;
        dfs(s, 0, temp, ret);
        return ret;
    }

    void dfs(string& s, int idx, vector<string>& temp, vector<vector<string>>& ret){
        if(idx == s.size()){
            ret.push_back(temp);
            return;
        }
        for(size_t i=1; i<=s.size()-idx; i++){
            string prefix = s.substr(idx, i);
            if(isPalindrome(prefix)){
                temp.push_back(prefix);
                dfs(s, idx+i, temp, ret);
                temp.pop_back();
            }
        }
    }
};