686. Repeated String Match

找substring存在不存在可以直接用indexOf() 或者contains() 不用一个一个比

边界情况是 c abc a 这样子 b.length()/a.length() == 1 所以要+2

 

class Solution {
    public int repeatedStringMatch(String A, String B) {
        if(B.length() == 0) return -1;
        int count = 1;
        String str = new String(A);
        for( count = 1; count <= B.length()/A.length() + 2; count++){
            if(str.contains(B) >= 0) return count;
            
            str = str + A;
        }
        return -1;
        
    }
}