[Haskell-beginners] Beginners Digest, Vol 45, Issue 35

[Ramesh Kumar]

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 at gmx.com" <franco00 at gmx.com>
>To: beginners at 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
>
>
>
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> 
>----- Original Message -----
>>From: franco00 at gmx.com
>>Sent: 03/28/12 09:36 AM
>>To: beginners at 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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