Java implementation LeetCode 686 repeat superimposed string matching

686. superimposing repeatedly a string matching

Given two strings A and B, to find the minimum number of repetitions of the A string is superimposed, so that the string become substrings A B after the addition, if there is no or -1.

For example, A = "abcd", B = "cdabcdab".

Answer is 3, because of "abcdabcdabcd", case B is the superposition of the sub strings A repeated three times; the A string is repeated twice superimposed "abcdabcd", B is not its children.

note:

A and B string length in the range of 1 and 10,000 interval.

class Solution {
   public int repeatedStringMatch(String A, String B) {
        
        
        //滚轮？ A作为滚轮
        char[] a = A.toCharArray();
        char[] b = B.toCharArray();
        for(int i=0;i<a.length;i++){
            int len = loop(a,b,i);
            if(len > 0){// 
                int count = 1;
                
                len = B.length() - a.length + i;
                count += len/a.length;
                count += len%a.length > 0 ? 1 : 0;
                return count;
            }else if(len + a.length <= 0){
                return -1;
            }
        }
        return -1;
        
    }
    //使用a滚轮印刷b，start为起始点
    public int loop(char[] a,char[] b,int start){
        int count = start;
        for(char c : b){
            if(a[start % a.length] != c){
                return count - start;
            }
            start++;
        }
        return 1; 
    }
}