Why don't you just swap the pattern match order?
remdups :: (Eq a) => [a] -> [a]
remdups (x : xx : xs) = if x == xx then remdups (x : xs) else x :
remdups (xx : xs)
remdups xs = xs
This should cover all cases no?
Also I prefer guards, but I guess that is personal
remdups (x1:x2:xs)
| x1 == x2 = remdups (x2 : xs)
| otherwise = x1 : remdups (x2 : xs)
remdups xs = xs
2009/3/15 R J
I need to write an implementation using foldl, and a separate implementation using foldr, of a function, "remdups xs", that removes adjacent duplicate items from the list xs. For example, remdups [1,2,2,3,3,3,1,1]= [1,2,3,1].
My approach is first to write a direct recursion, as follows:
remdups :: (Eq a) => [a] -> [a] remdups [] = [] remdups (x : []) = [x] remdups (x : xx : xs) = if x == xx then remdups (x : xs) else x : remdups (xx : xs)
This code works, but it has three cases, not usual two, namely [] and (x : xs).
What, if any, is the implementation using only two cases?
Also, if three cases are required, then how can it be implemented using foldr, and how using foldl?
Thanks.
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