2009/3/15 R J
What, if any, is the implementation using only two cases?
This version considers 2 cases 2 times :-) But only the 'go' function is recursive, so it could probably be written using some kind of fold. The value being build by the fold should probably be some kind of tuple so you can keep track of some state. remdups2 :: (Eq a) => [a] -> [a] remdups2 [] = [] remdups2 (x:xs) = go x xs where go x [] = [x] go x (y:ys) = if x == y then go x ys else x : go y ys I almost managed to write one using foldr, but it always put the first group last :-( Perhaps someone else manages to get it right. -- Warning: doesn't work correctly! Last group is last... remdups3 :: (Eq a) => [a] -> [a] remdups3 [] = [] remdups3 (x:xs) = snd $ foldr f (x, []) xs where f y (x, xs') = if y == x then (x, xs') else (y, x : xs')