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…

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…

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; : -…

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…

［抄题］: 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…

public static int repeatedStringMatch(String A, String B) { //判断字符串a重复几次可以包含另外一个字符串b,就是不断叠加字符串a直到长度大于等于b,叠加一次计数+1 //看现在的字符串是否包含b,包含就返回,不会包含就再叠加一次,因为可能有半截的在后边,再判断,再没有就返回-1 int count = 0; StringBuilder sb = new StringBuilder(); while (sb.length() < B.l…

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",B 并不是其子串. 注意: A 与 B 字符串的长度在1和10000区间范围内. 为什么(len2 /…

方法一.算是暴力解法吧,拼一段找一下 static int wing=[]() { std::ios::sync_with_stdio(false); cin.tie(NULL); ; }(); class Solution { public: int repeatedStringMatch(string A, string B) { string as=A; ; ;i<count;i++,as+=A) { if(as.find(B)!=string::npos) return i; } ; }…

题意 题目大意是,给两个字符串 A 和 B,问 B 是否能成为 A+A+A+...+A 的子字符串,如果能的话,那么最少需要多少个 A? 暴力解法 直接 A+A+...,到哪次 A 包含 B 了,就返回 A 的个数. 但是 B 也可能不是 A 的拼接的子字符串,所以这种直观解法还是存在隐患(无限循环),最好还是动动脑筋. 动脑筋解法 假如 B 的长度为 b,A 的长度为 a,那么 n=Math.ceil(b/a) 一定意味着什么.但是到底 n 意味着什么呢?看看例子先. 假设 A=“abcdef…