[Haskell-beginners] Longest common prefix of a list of lists
mokus at deepbondi.net
Fri Apr 29 21:42:17 CEST 2011
On Apr 29, 2011, at 10:32 AM, Daniel Fischer wrote:
> -- to get the common prefix of two lists, we use explicit recursion,
> don't see an elegant way to avoid that.
> commonPrefix :: Eq a => [a] -> [a] -> [a]
> commonPrefix (x:xs) (y:ys)
> | x == y = x : commonPrefix xs ys
> commonPrefix _ _ = 
> produces the result incrementally.
Personally, that's the definition I'd prefer. It's simple, clear, and
fast. If I were forced to hide the recursion, I would probably write
> commonPrefix a b = map fst (takeWhile (uncurry (==)) (zip a b))
modulo usage of whatever combination of (.) and/or ($) is in style
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