通常情况下,我们判断一个字符串中是否存在某值常常会用string.contains,其实判断一个字符串中存在某值的方法有很多种,最常用的就是前述所说的string.contains,相对来说比较常用的还有string.IndexOf和Regex.Match.直接上代码,后面在说些什么吧,通常情况下功能的实现最重要,作者的话,只对有心者有效. using System; using System.Collections.Generic; using System.Linq; using Syste

string.match(RegExp) 与 RegExp.exec(string) 相同点与不同点对比解析: 1. 这两个方法,如果匹配成功,返回一个数组,匹配失败,返回null. 2. 当RegExp的global属性为false时,这两个方法的返回数组是一样的. 数组的第0个元素是整个str的第一个匹配字符串,接下来的元素是str第一个匹配中的子匹配字符串. 此外,数组还有index和input两个额外属性,index是匹配字符串的起始位置,input是整个输入字符串. 此时,RegExp

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

942. 增减字符串匹配 942. DI String Match 题目描述 每日一算法2019/6/21Day 49LeetCode942. DI String Match Java 实现 and so on 参考资料 https://leetcode-cn.com/problems/di-string-match/ https://leetcode.com/problems/di-string-match/

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",

problem 942. DI String Match 参考 1. Leetcode_easy_942. DI String Match; 完

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; : -

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

今天在重新阅读<JavaScript权威指南>的RegExp和String的时候,看到了2个比较容易混淆的函数:RegExp的exec和String的match 这2个函数都是从指定的字符串中提取符合条件的字串.他们大部分时候的返回结构是一致的,但是在全局检索时,返回的结果差别会很大1.正则表达式为非全局检索时,2者是等价的 返回符合条件的第一个子串以及对应的分组(如果存在存在)构成的伪数组:第一个元素是匹配的字串,余下的为匹配的分组. 伪数组包含2个属性: input 返回要检索的字符串 i

#include <stdio.h> #include <string.h> #include <stdlib.h> void SubString(char sub[], char s[], int i, int m) { int j; ; j<=m; j++) sub[j]=s[i++]; sub[j]=NULL; } int main() { ], c[], *sub=NULL; int num,m,n,i,count; scanf("%d"

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

Given a string S that only contains "I" (increase) or "D" (decrease), let N = S.length. Return any permutation A of [0, 1, ..., N] such that for all i = 0, ..., N-1: If S[i] == "I", then A[i] < A[i+1] If S[i] == "D&qu

这是悦乐书的第289次更新,第307篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第156题(顺位题号是686).给定两个字符串A和B,找到A必须重复的最小次数,使得B是它的子字符串. 如果没有这样的解决方案,返回-1.例如: 输入:A ="abcd",B ="cdabcdab". 输出:3 说明:因为重复A三次("abcdabcdabcd"),B是它的子串; 和B不是A重复两次的子串("abcdabcd&

Finding length of longest common substring /*Finding length of longest common substring using DP * */ import java.util.*; public class Solution { /* * Returns length of longest common substring of * X[0...m-1] and Y[0...n-1] * */ public static int LC

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

https://leetcode.com/problems/di-string-match/ Given a string S that only contains "I" (increase) or "D" (decrease), let N = S.length. Return any permutation A of [0, 1, ..., N] such that for all i = 0, ..., N-1: If S[i] == "I&quo

题目要求 Given a string S that only contains "I" (increase) or "D" (decrease), let N = S.length. Return any permutation A of [0, 1, ..., N] such that for all i = 0, ..., N-1: If S[i] == "I", then A[i] < A[i+1] If S[i] == "

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 substring of it. If no such solution, return -1. For example, with A = "abcd" and B = "cdabcdab". Return 3, because by repeating A

［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