Rabin Karp字符串匹配算法复杂度

Question

在rabin算法中取模如何帮助降低原生horners规则字符串匹配的复杂性。

Answer 1

I guess by Horners rule you mean treating a string as a number in some base ("abcd" = 'a' * p^3 + 'b' * p^2 + 'c' * p^1 + 'd' * p^0) and comparing strings as numbers, and by Rabin algorithm you mean the essentially the same thing but modulo some other number.

The thing is using Horners rule you can only compare short strings -- otherwise you get overflow (you can use big integers to avoid it, but that is where you lose in complexity. The number corresponding to a string of length n will have O(n) digits, so arithmetic operations won't be done in O(1)).

And in Rabin-Karp algorithm the numbers corresponding to our strings will stay small since we take them modulo some other number. It may cause collisions, but if we are lucky, collisions are rare enough.