686. Repeated String Match 重复字符串匹配

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.


给定两个字符串A和B，找到必须重复的最小次数，使得B是其子串。如果没有这样的解决方案，返回-1。


  1. /**
  2.  * @param {string} A
  3.  * @param {string} B
  4.  * @return {number}
  5.  */
  6. var repeatedStringMatch = function (A, B) {
  7.     if (A.includes(B)) {
  8.         return 1;
  9.     }
  10.     let time = 1;
  11.     while (true) {
  12.         let res = A.repeat(++time);
  13.         if (res.includes(B)) {
  14.             return time;
  15.         }else if (res.length > B.length) {
  16.             return -1;
  17.         }
  18.     }
  19. };







posted @ 2017-10-04 23:09  xiejunzhao  阅读(206)  评论(0)    收藏  举报