原题链接在这里：https://leetcode.com/problems/repeated-substring-pattern/
题目：
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.)
题解：
若是str能被拆能n个重复的子串, 那么每个子串的长度不会超过str.length()/2. 从str.length()/2到1长度，挨个试取字串拼回去，看是否等于原串.
Time Complexity: O(n^2). 拆成n/2, 拼回去用时2, 拆成n/3, 拼回去用时3, ..., 总用时(2+3+4+...+n), n = str.length().
Space: O(1).
AC Java:
1 public class Solution { 2 public boolean repeatedSubstringPattern(String str) { 3 int len = str.length(); 4 for(int i = len/2; i>=1; i--){ 5 if(len%i == 0){ 6 String template = str.substring(0, i); 7 StringBuilder sb = new StringBuilder(); 8 int n = len/i; 9 while(n-->0){ 10 sb.append(template); 11 } 12 if(sb.toString().equals(str)){ 13 return true; 14 } 15 } 16 } 17 return false; 18 } 19 }
可采用类似KMP的算法. 这里的next不再是当前char之前的最大相同前缀后缀，而是包含当前char的最大相同前缀后缀. e.g "abcabcabc", next = [0,0,0,1,2,3,4,5,6].
看str.length()能否被str.length() - next[next.length-1]整除.
Time Complexity: O(str.length()). Space: O(str.length()).
AC Java:
1 public class Solution { 2 public boolean repeatedSubstringPattern(String str) { 3 if(str == null || str.length() == 0){ 4 return true; 5 } 6 int len = str.length(); 7 int [] next = new int[len]; 8 getNext(str, next); 9 int preLen = next[next.length-1]; 10 return (preLen>0 && len%(len-preLen) == 0); 11 } 12 13 private void getNext(String s, int [] next){ 14 int len = s.length(); 15 int p = 0; 16 int q = 1; 17 while(q < len){ 18 //s.charAt(p)表示前缀. s.charAt(q)表示后缀 19 if(s.charAt(p) == s.charAt(q)){ 20 next[q] = p+1; 21 p++; 22 q++; 23 }else if(p == 0){ 24 next[q] = p; 25 q++; 26 }else{ 27 p = next[p-1]; 28 } 29 } 30 } 31 }