686. Repeated String Match
Easy
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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.
注意find()的应用,另外while(C.length()<B.length())这里最开始我写反了,写成>,调了一会儿才发现。
class Solution {
public:
int repeatedStringMatch(string A, string B) {
string C="";
int ans=0;
while(C.length()<B.length())
{
C+=A;
ans++;
}
if(C.find(B)!=-1)
{
return ans;
}
// else
// {
C+=A;
ans++;
if(C.find(B)!=-1)
{
return ans;
}
else return -1;
// }
}
};