LWC 52：686. Repeated String Match

LWC 52：686. Repeated String Match

传送门：686. Repeated String Match

Problem:

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.

思路：
重复多少次之后能够使得A包含B，关键在于何时停止重复，显然，如果A的repeat后的长度大于B时，即可停止搜索了，因为在此长度下A都不能包含B，那么repat的次数再大也没用。

代码如下：

public int repeatedStringMatch(String A, String B) {
        int nb = B.length();
        int na = A.length();
        int times = nb / na + 2;
        StringBuilder sb = new StringBuilder(A);
        for (int i = 1; i <= times; ++i) {
            if (sb.toString().contains(B)) return i;
            else {
                sb.append(A);
            }
        }
        return -1;
    }

times的上界可以设置的大点，当然+2已经是最紧的上界了。