686. Repeated String Match

最新推荐文章于 2023-10-14 20:31:28 发布

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最新推荐文章于 2023-10-14 20:31:28 发布

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原文链接：http://www.cnblogs.com/jessie2009/p/9817922.html

本文介绍了一个算法，用于解决给定两个字符串A和B，寻找A最少需要重复多少次才能使B成为其子串的问题。通过StringBuilder动态增加A的长度直至包含B，最终返回所需重复次数。若无法实现则返回-1。

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.

//Time: O(n), Space: O(1)   
 public int repeatedStringMatch(String A, String B) {
        if (A == null || B == null) {
            return 0;
        }
        
        int count = 1;
        StringBuilder sb = new StringBuilder(A);
        
        while (sb.length() < B.length()) {
            sb.append(A);
            count++;
        }
        
        if (sb.toString().contains(B)) {
            return count;
        }
        
        if (sb.append(A).toString().contains(B)) {
            return count + 1;
        }
        return -1;
    }