[Haskell-cafe] Re: Subsequence near solved hopefully
Sam Mason
mason at f2s.com
Sun Oct 17 17:53:37 EDT 2004
Peter Simons wrote:
>This version should do it:
>
>isSubSeq :: (Eq a) => [a] -> [a] -> Bool
>isSubSeq [] _ = True
>isSubSeq _ [] = False
>isSubSeq (x:xs) (y:ys)
> | x == y = isSubSeq xs ys
^^^^^^^^
I think you want to refer to List.isPrefixOf here - your version is a
sort of "ordered subset" test. I.e. I get:
"abc" `isSubSeq` ".a.b.c." ===> True
> | otherwise = isSubSeq (x:xs) ys
My version would've been:
isSubSeq x = any (isPrefixOf x) . tails
But Remi beat me to it (and for that I'll never forgive him! :-).
Sam
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