686. 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.

重复几次A后，B是A的子串

C++（16ms）：

class Solution {
public:
    int repeatedStringMatch(string A, string B) {
        string t = A ;
        for(int i = 1 ; i <= B.length()/A.length()+2 ; i++){
            if (t.find(B) != string::npos){
                return i ;
            }else{
                t += A ;
            }
        }
        return -1; 
    }
};

Java(412ms):

class Solution {
    public int repeatedStringMatch(String A, String B) {
        String t = A ;
        for(int i = 1 ; i <= B.length()/A.length()+2 ; i++){
            if (t.indexOf(B) != -1)
                return i ;
            else
                t += A ;
        }
        return -1 ;
    }
}