Thanks Franco, Your (first) solution is the only one which has worked so far although it utilizes a lambda expression.
The problem is indeed tricky.




From: "franco00@gmx.com" <franco00@gmx.com>
To: beginners@haskell.org
Sent: Wednesday, March 28, 2012 3:39 PM
Subject: Re: [Haskell-beginners] Beginners Digest, Vol 45, Issue 35

gah sorry I obviously meant to reply to the "Unique integers in a list" message



 
----- Original Message -----
From: franco00@gmx.com
Sent: 03/28/12 09:36 AM
To: beginners@haskell.org
Subject: Re: Beginners Digest, Vol 45, Issue 35

unique :: [Integer] -> [Integer]
unique []   = []
unique (x:xs) | elem x xs   = (unique . filter (/= x)) xs
              | otherwise   = x : unique xs

-- This is a simpler to read version (albeit inefficient?)
unique :: [Integer] -> [Integer]
unique []   = []
unique (x:xs) | elem x xs   = unique xs
              | otherwise   = x : unique xs
 

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