Re: [Haskell-beginners] Longest common prefix of a list of lists

On Apr 29, 2011, at 10:32 AM, Daniel Fischer wrote:

-- to get the common prefix of two lists, we use explicit recursion, I 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 this:

commonPrefix a b = map fst (takeWhile (uncurry (==)) (zip a b))

modulo usage of whatever combination of (.) and/or ($) is in style today. -- James