LeetCode: Palindrome Partitioning

思路：很显然的深度搜索，如果当前子串是回文串，继续从子串末尾的下一个字符开始搜索。

code:

class Solution {
public:
    bool  isPalindrome(string s, int n){
        if(n == 1)return true;
        int l = 0, r = n - 1;
        while(l < r)
            if(s[l++] != s[r--])
                return false;
        return true;
    }
    
    void dfs(int start,int n,vector<vector<string>> &ret, vector<string> &curStr, string s){
        if(start == n){
            ret.push_back(curStr);
            return;
        }
        for(int i = start;i < n;i++){
            string temp = s.substr(start,i-start+1);
            if(isPalindrome(temp,i-start+1)){
                vector<string> str(curStr);
                str.push_back(temp);
                dfs(i+1,n,ret,str,s);
            }
        }
    }
    
    vector<vector<string>> partition(string s) {
        vector<vector<string>> ret;
        int n = s.length();
        vector<string> curStr;
        dfs(0,n,ret,curStr,s);
        return ret;
    }
};