问题
So my question is in topic's name. Does exists an algorithm that checks if B is subsequence of A faster, than O(N^2), for example O(NlogN) or simply O(N)?
Only way found is simple brut-force
for(int i = 0; i < a.Length - b.Length; i++)
{
if (IsSubsequence(a,b,i))
return i;
}
return -1;
回答1:
Here's a recursive characterization of David Eisenstat's algorithm. (Note that this algorithm is tail recursive and can therefore be written as a loop; I describe it as recursive because doing so is a nice way to understand the algorithm.)
Define a sequence as either empty, or an item followed by a sequence.
Take two sequences, A and B. The question is if B is or is not a subsequence of A.
If B is empty then B is a subsequence of A.
If B is not empty and A is empty then B is not a subsequence of A.
If we've made it this far, neither A nor B are empty. Suppose that A is item X followed by sequence C and B is item Y followed by sequence D.
If X is the same as Y then the answer to the question "is B a subsequence of A?" is the same as the answer to the smaller question "is D a subsequence of C?". Answer that question.
If X is not the same as Y then the answer to the question "is B a subsequence of A?" is the same as the answer to the smaller question "is B a subsequence of C?". Answer that question.
This procedure terminates and clearly its worst case is in the length of the sequence A.
回答2:
Yes, it can be done faster than O(n^2) with an algorithm by Knuth, Morris and Pratt. However note that the implementation provided by the framework might already implement this algorithm.
回答3:
Yes, the following greedy algorithm is correct and runs in time O(n). For each element of B in order, scan forward in A from the previous stopping point (initially the beginning of A) for the first occurrence.
来源：https://stackoverflow.com/questions/22637634/can-i-check-if-subsequence-faster-then-onn