686. Repeated String Match - Easy

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.

 

M1: brute force

keep string builder and appending until the length A is greater or equal to B
B要能成为A的字符串，那么A的长度肯定要大于等于B，所以当A的长度小于B的时候，先重复A，直到A的长度大于等于B，并且累计次数cnt。此时找B是否存在A中，如果存在直接返回cnt。如果不存在，加上一个A再找（这样可以处理这种情况A="abc", B="cab"）如果此时还找不到，说明无法匹配，返回-1。

时间：O(M+N)，空间：O(max(M, N))

class Solution {
    public int repeatedStringMatch(String A, String B) {
        StringBuilder sb = new StringBuilder();
        sb.append(A);
        int cnt = 1;
        while(sb.length() < B.length()) {
            sb.append(A);
            cnt++;
        }
        if(sb.toString().contains(B))
            return cnt;
        else if(sb.append(A).toString().contains(B))
            return cnt+1;
        else
            return -1;
    }
}

 

M2: KMP

时间：O(N)