686. Repeated String Match

686. Repeated String Match

• 题目描述：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”).

• 题目大意：给定两个字符串A,B，判断A重复多少次，B会变为其的子串，如果无解，返回-1。

• 思路：使用StringBulider

• 代码

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