Given two strings A and B, find the minimum number of times A has to be repeated such that B is a substring of it. If no such solution, return -1. For example, with A = "abcd" and B = "cdabcdab". Return 3, because by repeating A three…
Given two strings A and B, find the minimum number of times A has to be repeated such that B is a substring of it. If no such solution, return -1. For example, with A = "abcd" and B = "cdabcdab". Return 3, because by repeating A three…
原题链接在这里:https://leetcode.com/problems/repeated-string-match/description/ 题目: Given two strings A and B, find the minimum number of times A has to be repeated such that B is a substring of it. If no such solution, return -1. For example, with A = "abc…
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.com/problems/repeated-string-match/description/ 题目描述 Given two strings A and B, find the minimum number of times A has to be repeated such that B is a…
686. 重复叠加字符串匹配 686. Repeated String Match 题目描述 给定两个字符串 A 和 B,寻找重复叠加字符串 A 的最小次数,使得字符串 B 成为叠加后的字符串 A 的子串,如果不存在则返回 -1. 举个例子,A = "abcd",B = "cdabcdab". 答案为 3,因为 A 重复叠加三遍后为 "abcdabcdabcd",此时 B 是其子串:A 重复叠加两遍后为 "abcdabcd",…
题目:Given two strings A and B, find the minimum number of times A has to be repeated such that B is a substring of it. If no such solution, return -1. For example, with A = "abcd" and B = "cdabcdab". Return 3, because by repeating A thr…
problem 686. Repeated String Match solution1: 使用string类的find函数: class Solution { public: int repeatedStringMatch(string A, string B) { ; string t = A; while(t.size() < n2) { t += A; cnt++; } if(t.find(B) != string::npos) return cnt;//err. t += A; : -…
