Difference between revisions of "99 questions/Solutions/14"

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Line 35: Line 35:
 
<haskell>
 
<haskell>
 
dupli = foldr (\ x xs -> x : x : xs) []
 
dupli = foldr (\ x xs -> x : x : xs) []
 +
</haskell>
 +
 +
or, using silliness:
 +
<haskell>
 +
dupli = foldr (\x -> ((x:) . (x:)) []
 
</haskell>
 
</haskell>

Revision as of 21:00, 13 January 2011

(*) Duplicate the elements of a list.

dupli [] = []
dupli (x:xs) = x:x:dupli xs

or, using list comprehension syntax:

dupli list = concat [[x,x] | x <- list]

or, using the list monad:

dupli xs = xs >>= (\x -> [x,x])

or, using concatMap:

dupli = concatMap (\x -> [x,x])

also using concatMap:

dupli = concatMap (replicate 2)

or, using foldl:

dupli = foldl (\acc x -> acc ++ [x,x]) []

or, using foldr:

dupli = foldr (\ x xs -> x : x : xs) []

or, using silliness:

dupli = foldr (\x -> ((x:) . (x:)) []