It can be, but your way traverses the entire haystack *twice*. The memory
needs are equivalent to the reversing version, which I consider
unacceptable.
On Sat, Oct 11, 2014 at 5:57 AM, Andreas Abel
David, the essence of your idea seems mutually drop the correct number of elements from needle and hay and then compare for equality. Here is a concise implementation of your idea in terms of drop:
isSuffixOf :: forall a . (Eq a) => [a] -> [a] -> Bool [] `isSuffixOf` _ = True xs `isSuffixOf` ys = xs == drop (length (drop (length xs) ys)) ys
This can be further simplified to
isSuffixOf :: forall a . (Eq a) => [a] -> [a] -> Bool [] `isSuffixOf` _ = True xs `isSuffixOf` ys = xs == drop (length ys - length xs) ys
which is a very easy to understand program, I think, without need to reverse lists.