AI代码解释 intrepeatedStringMatch(stringA,stringB){string newA;int count=0;while(newA.size()<B.size())//重复A使得newA的长度大于等于B{newA+=A;count++;//记录重复的次数}if(newA.find(B)!=-1)//如果找得到,返回重复次数returncount;newA+=A;//如果找不到,再重复一次Aif(newA.find(B)!=-1)...
LeetCode686——Repeated String Match 1 2 3 4 5 6 7 8 题目:<br>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 ...
classSolution {public:intrepeatedStringMatch(stringA,stringB) {intn1 = A.size(), n2 = B.size(), cnt =1;stringt =A;while(t.size() <n2) { t+=A;++cnt; }if(t.find(B) !=string::npos)returncnt; t+=A;return(t.find(B) !=string::npos) ? cnt +1: -1; } }; 下面这种解法...
Can you solve this real interview question? Repeated String Match - Given two strings a and b, return the minimum number of times you should repeat string a so that string b is a substring of it. If it is impossible for b to be a substring of a after rep
int repeatedStringMatch(string a, string b) { int count = 1; string aa = a; while (aa.length() < b.length()) { count++; aa += a; } if (aa.find(b) != aa.npos) return count; else { aa += a; if (aa.find(b) != aa.npos) ...
class Solution: def repeatedStringMatch(self, a: str, b: str) -> int: if not set(b).issubset(set(a)): return -1 cnt = 1 while len(a * cnt) < 2 * len(a) + len(b): if b in a * cnt: return cnt cnt += 1 return -1 复杂度分析 时间复杂度：...
public:intrepeatedStringMatch(stringA,stringB) {stringt =A;for(inti =1; i <= B.size() / A.size() +2; ++i) {if(t.find(B) !=string::npos)returni; t+=A; }return-1; } }; 下面这种解法还是由热心网友edyyy提供,没有用到字符串自带的find函数,而是逐个字符进行比较,循环字符串A中的...
def repeatedStringMatch(self, a: str, b: str) -> int: # # 思路1、根据字符串长度关系分析出 重复次数的最小取值和最大取值 # result = 1 # # 重复次数的最小取值:保证长度上能够覆盖b字符串 # if len(b) % len(a) == 0: # result = len(b) // len(a) # else: # result = len(...
Leetcode: Repeated String Match 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"....
public int repeatedStringMatch(String a, String b) { int count = b.length() / a.length() + 2;for (int i = 1; i <= count; i++) { StringBuilder sb = new StringBuilder();for (int j = 0; j i; j++) { sb.append(a);} String repeatedA = sb.toString();if (repeatedA....