https://leetcode.com/problems/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".
Return 3, because by repeating A three times (“abcdabcdabcd”), B is a substring of it; and B is not a substring of A repeated two times ("abcdabcd").
Note:
The length of A and B will be between 1 and 10000.
代码：
class Solution {
public:
int repeatedStringMatch(string A, string B) {
int la = A.length(), lb = B.length();
map<char, int> mp;
for(int i = 0; i < la; i ++) {
if(mp[A[i]] == 0) mp[A[i]] ++;
}
for(int i = 0; i < lb; i ++) {
if(mp[B[i]] == 0) return -1;
}
if(A.find(B) != -1) return 1;
string c = A;
for(int i = 1; i <= lb; i ++) {
c += A;
if(c.find(B) != -1) return i + 1;
if(c.length() > B.length() * 3) return -1;
}
return 0;
}
};
又来？行吧行吧 bug 都是自己的锅 今天的 LeetCode 五题打卡
也许一会应该醒了
#Leetcode# 686. Repeated String Match
原文：https://www.cnblogs.com/zlrrrr/p/10692326.html