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…
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…
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…
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…
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…
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…
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…
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/…
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…
#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 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…
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…
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] == "…
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…
题目如下: 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] == &quo…
