459. Repeated Substring Pattern

QUESTION:

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.)

EXPLANATION:

首先重复的肯定是大于等于两个的，所以可以从中间开始算作能够重复的最大字符串，然后再逐步递减。如果无法整除的话肯定就是无法重复的，所以在整除的基础上，将截取的字符串*对应的倍数生成整个字符串，再与元字符串进行比对。如果循环都走完了，说明没有可以重复的字符串。

SOLUTION:

    public boolean repeatedSubstringPattern(String str) {
        int n  = str.length();
        for(int i =n/2;i>=1;i--){
            if(n%i==0){
                String substring = str.substring(0,i);
                StringBuilder sb = new StringBuilder();
                for(int j = 0;j<n/i;j++){
                    sb.append(substring);
                }
                if(sb.toString().equals(str)) return true;
            }
        }
        return false;
    }
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