新手问题 最近做 InterviewStreet.com 上的题, String Similarity 卡住了

Kineo_88 · 2012年10月11日 · 最后由 Kineo_88 回复于 2012年10月11日 · 2810 次阅读

题目是这样:

For two strings A and B, we define the similarity of the strings to be the length of the longest prefix common to both strings. For example, the similarity of strings "abc" and "abd" is 2, while the similarity of strings "aaa" and "aaab" is 3.

Calculate the sum of similarities of a string S with each of it's suffixes.

Input: The first line contains the number of test cases T. Each of the next T lines contains a string each.

Output: Output T lines containing the answer for the corresponding test case.

Constraints: 1 <= T <= 10 The length of each string is at most 100000 and contains only lower case characters.

Sample Input: 2 ababaa aa

Sample Output: 11 3

Explanation: For the first case, the suffixes of the string are "ababaa", "babaa", "abaa", "baa", "aa" and "a". The similarities of each of these strings with the string "ababaa" are 6,0,3,0,1,1 respectively. Thus the answer is 6 + 0 + 3 + 0 + 1 + 1 = 11.

For the second case, the answer is 2 + 1 = 3.

用 C 写了一个,但是换到 Ruby 就完全卡住了...囧 希望各位大牛不吝赐教啊!

扩展 kmp 算法。google 一下就知道了,复杂度 O(length).

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