LWC 52:686. Repeated String Match
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LWC 52:686. Repeated String Match
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Problem:
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.
思路:
重复多少次之后能够使得A包含B,关键在于何时停止重复,显然,如果A的repeat后的长度大于B时,即可停止搜索了,因为在此长度下A都不能包含B,那么repat的次数再大也没用。
代码如下:
public int repeatedStringMatch(String A, String B) { int nb = B.length(); int na = A.length(); int times = nb / na + 2; StringBuilder sb = new StringBuilder(A); for (int i = 1; i <= times; ++i) { if (sb.toString().contains(B)) return i; else { sb.append(A); } } return -1; }
times的上界可以设置的大点,当然+2已经是最紧的上界了。
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