Can you solve this real interview question? Repeated Substring Pattern - Given a string s, check if it can be constructed by taking a substring of it and appending multiple copies of the substring together. Example 1: Input: s = "abab" Output: true
publicclassSolution { publicbooleanrepeatedSubstringPattern(String str) { intn = str.length(); for(inti=n/2;i>=1;i--) { if(n%i==0) { intm = n/i; String substring = str.substring(0,i); StringBuilder sb =newStringBuilder(); for(intj=0;j<m;j++) { sb.append(substring); } if...
leetcode 459. Repeated Substring Pattern 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....
leetcode 459. Repeated Substring Pattern 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....
}returndp[n] && (dp[n] % (n - dp[n]) ==0); } }; 本文转自博客园Grandyang的博客,原文链接:重复子字符串模式[LeetCode] Repeated Substring Pattern,如需转载请自行联系原博主。
LeetCode Repeated Substring Pattern 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 ...
Also, for each 'left' and 'right', start expanding left-- and right-- as long as they match the pattern of palindromic. Whenever failed to match, return whatever length of palindromic substring. Track the max length and max string. ...
Explanation:It'sthe substring"ab"twice. Example 2: Input:"aba"Output:False Example 3: Input:"abcabcabcabc"Output:True Explanation:It'sthe substring"abc"four times.(And the substring"abcabc"twice.) 一开始是没有思路的。。 defrepeatedSubstringPattern2(self,str):""" ...
publicbooleanrepeatedSubstringPattern4(Strings) {Stringstr = s + s;returnstr.substring(1, str.length() -1).contains(s); } 06 小结 算法专题目前已日更超过三个月,算法题文章103+篇,公众号对话框回复【数据结构与算法】、【算法】、【数据结构】中的任一关键词,获取系列文章合集。
classSolution {public:boolrepeatedSubstringPattern(stringstr) {inti =1, j =0, n =str.size(); vector<int> dp(n +1,0);while(i <n) {if(str[i] == str[j]) dp[++i] = ++j;elseif(j ==0) ++i;elsej =dp[j]; }returndp[n] && (dp[n] % (n - dp[n]) ==0); ...