标签：inter 直接 sts amp 判断 div bsp only multi
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.)
题目含义：判断一个字符串，能够由其子串的多个拷贝
1 public boolean repeatedSubstringPattern(String s) { 2 int l = s.length(); 3 for (int i = l / 2; i >= 1; i--) { 4 if (l % i != 0) continue; //l为奇数的，一定不满足条件,直接跳过 5 int m = l / i; //拷贝的数量 6 String item = s.substring(0, i);//依次判断从0到i组成的子串 7 StringBuilder sb = new StringBuilder(); 8 while (m > 0) { 9 sb.append(item); 10 m--; 11 } 12 //判断m个子串能否构成主串 13 if (sb.toString().equals(s)) return true; 14 } 15 return false; 16 }
459. Repeated Substring Pattern
标签：inter 直接 sts amp 判断 div bsp only multi
原文地址：http://www.cnblogs.com/wzj4858/p/7680262.html
