459. 重复的子字符串

为保证权益，题目请参考 459. 重复的子字符串(From LeetCode).

解决方案1

CPP

//
// Created by lenovo on 2020-08-24.
//

#include <iostream>

using namespace std;

class Solution {
public:
    bool repeatedSubstringPattern(string s) {
        int n = s.length();
        for (int i = 1; i < n; i++) {
            if (n % i == 0) {
                string t = s.substr(0, i);
                if(isit(s, t)){
                    return true;
                }
            }
        }
        return false;
    }

    /**
     * 判断t是不是s的循环子串
     *
     * @param s
     * @param t
     * @return
     */
    bool isit(const string & s, const string & t){
        int sn = s.length();
        int tn = t.length();

        for(int i = 0; i < sn; i += tn){
            if(s.substr(i, tn) != t) {
                return false;
            }
        }
        return true;
    }
};

int main() {
    return 0;
}

Python

"""
LeetCode 459

https://leetcode-cn.com/problems/repeated-substring-pattern/
https://leetcode-cn.com/problems/repeated-substring-pattern/
https://leetcode-cn.com/problems/repeated-substring-pattern/

"""


class Solution:
    """
    静态类
    """

    def repeatedSubstringPattern(self, s: str) -> bool:
        """
        题目要求

        :param s:
        :return:
        """
        for i in range(1, len(s), 1):
            if len(s) % i == 0:
                t = s[0:i]
                if self.isit(s, t):
                    return True
        return False

    def isit(self, s: str, t: str) -> bool:
        """
        判断t是否是S的循环子串


        :param s: 大串
        :param t: 小串
        :return: bool
        """
        for i in range(0, len(s), len(t)):
            if s[i:i + len(t)] != t:
                return False
        return True