LeetCode Repeated String Match

原题链接在这里：https://leetcode.com/problems/repeated-string-match/description/

题目：

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.

题解：

A重复append自己知道比B长，看B是否包含在内，若包含返回当前count.

若没有再append次A. 若B能是substring, 这个长度已经可以包含所有可能性. B的开头在0到A.length()-1的任何位置都足够盖住B.

Time Complexity: O(A.length()+B.length()). create个长度为A.length()+B.length()的string. 再用B找index.

Space: O(A.length()+B.length()).

AC Java:

class Solution {
    public int repeatedStringMatch(String A, String B) {
        int count = 0;
        StringBuilder sb = new StringBuilder();
        while(sb.length() < B.length()){
            sb.append(A);
            count++;
        }
        
        if(sb.indexOf(B) >= 0){
            return count;
        }else if(sb.append(A).indexOf(B) >= 0){
            return count+1;
        }else{
            return -1;
        }
    }
}

LeetCode Repeated String Match

原文：http://www.cnblogs.com/Dylan-Java-NYC/p/7684884.html