459. Repeated Substring Pattern

原创于 2020-01-01 19:16:35 发布 · 228 阅读

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本文探讨了如何通过算法判断一个非空字符串是否能够由其自身的一个子串重复拼接而成。提供了Java和Python实现代码，通过遍历并检查子串重复构成的新串与原串是否相等来解决问题。

459. Repeated Substring Pattern

Easy

Given a non-empty string check if it can be constructed by taking a substring of it and appending multiple copies of the substring together. You may assume the given string consists of lowercase English letters only and its length will not exceed 10000.

Example 1:

Input: "abab"
Output: True
Explanation: It's the substring "ab" twice.

Example 2:

Input: "aba"
Output: False

Example 3:

Input: "abcabcabcabc"
Output: True
Explanation: It's the substring "abc" four times. (And the substring "abcabc" twice.)

这道题的题意是让我们判断一个字符串是否能拆成多个子串，既然能拆成多个子串，那么每个子串的长度肯定不能大于字符串的一半。

从字符串的一半开始遍历
子串从 0 -> i 截取，然后将子串重复 len / i 次，构成新串
判断和原来的字符串是否相等

Java

class Solution {
    public boolean repeatedSubstringPattern(String s) {
        int len = s.length();
        for(int i = 1; i <= len/2; i++){
            String sub = s.substring(0, i);
            StringBuffer sb = new StringBuffer();
            for(int j=0; j<len/i; j++){
                sb.append(sub);
            }
            if(sb.toString().equals(s)){
                return true;
            }
        }
        return false;
    }
}

Python

class Solution:
    def repeatedSubstringPattern(self, s: str) -> bool:
        len_ = len(s)
        for i in range(1, len_//2+1):
            if len_ % i == 0:
                sub_str = s[:i]
                if sub_str * (len_//i) == s:
                    return True
        return False