686. Repeated String Match 重复字符串匹配

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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. let isLonger = false;
  12. while (isLonger == false) {
  13. let res = A.repeat(++time);
  14. if (res.includes(B)) {
  15. return time;
  16. }
  17. if (res.length > B.length) {
  18. isLonger = true;
  19. }
  20. }
  21. return -1;
  22. };






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