【问题描述】[中等]

【解答思路】

1. 枚举

找出能整除的子串长度，再用substring遍历匹配即可
时间复杂度：O(N^2) 空间复杂度：O(1)

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);int lenSub = sub.length();int count =0 ;//统计字段数 正确的话应该有 d-1int d = 0; //一共有多少段if(len%lenSub ==0){d = len/lenSub -1;for(int  j=i;j+lenSub<=len;j+=lenSub){String sub2 = s.substring(j,j+lenSub);if(sub.equals(sub2)){count++;}}if(count == d){return true;}}}return false;}
}

优化 子串 flag

class Solution {public boolean repeatedSubstringPattern(String s) {for(int i = 1; i < s.length(); ++i){if(s.length()%i == 0){String t = s.substring(0, i);boolean flag = true;for(int j = i; j + i <= s.length(); j += i){if(!t.equals(s.substring(j, j + i))){flag = false;break;}}if(flag) return true;}}return false;}
}

优化 逐个比较

class Solution {public boolean repeatedSubstringPattern(String s) {int n = s.length();for (int i = 1; i * 2 <= n; ++i) {if (n % i == 0) {boolean match = true;for (int j = i; j < n; ++j) {if (s.charAt(j) != s.charAt(j - i)) {match = false;break;}}if (match) {return true;}}}return false;}
}

2. 字符串匹配

class Solution {public boolean repeatedSubstringPattern(String s) {return (s + s).indexOf(s, 1) != s.length();}
}

【总结】

1. 扩展思路 KMP算法

2.思路二正确性证明

3.优化细节

3.1 子串长度至多等于原字符串除以2
3.2 flag标志+ break提前终止

转载链接：https://leetcode-cn.com/problems/repeated-substring-pattern/solution/zhong-fu-de-zi-zi-fu-chuan-by-leetcode-solution/
参考链接：https://leetcode-cn.com/problems/repeated-substring-pattern/solution/zhong-fu-de-zi-zi-fu-chuan-by-leetcode-solution/