131. Palindrome Partitioning

Problem statement:

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"]]

Solution:
This problem returns all possible palindrome partitions. Obviously, it is DFS.

class Solution {
public:
    vector<vector<string>> partition(string s) {
        vector<vector<string>> partition_set; 
        vector<string> partition;
        partition_dfs(partition_set, partition, s, 0);
        return partition_set;
    }
private:
    void partition_dfs(vector<vector<string>>& partition_set, vector<string>& partition, string& s, int start){
        if(start == s.size()){
            partition_set.push_back(partition);
            return;
        }
        for(int i = start + 1; i <= s.size(); i++){
            string str = s.substr(start, i - start);
            if(is_parlindrome(str)){
                partition.push_back(str);
                partition_dfs(partition_set, partition, s, i);
                partition.pop_back();
            }
        }
        return;
    }
    bool is_parlindrome(string s){
        int left = 0; 
        int right = s.size() - 1;
        while(left < right){
            if(s[left] != s[right]){
                return false;
            }
            left++;
            right--;
        }
        return true;
    }
};