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99 questions/Solutions/15

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(discovered an interesting solution that was not on here)
Current revision (16:55, 19 November 2012) (edit) (undo)
m (Improved my version a little)
 
Line 32: Line 32:
</haskell>
</haskell>
-
or, a convoluted recursive solution that only uses cons:
+
or, a version that does not use list concatenation:
<haskell>
<haskell>
repli :: [a] -> Int -> [a]
repli :: [a] -> Int -> [a]
repli [] _ = []
repli [] _ = []
-
repli (x:xs) n = foldl (\f _ -> (x:) . f) (repli xs) [1..n] n
+
repli (x:xs) n = foldr (const (x:)) (repli xs n) [1..n]
</haskell>
</haskell>

Current revision

(**) Replicate the elements of a list a given number of times. 

repli :: [a] -> Int -> [a]
repli xs n = concatMap (replicate n) xs

or, in Pointfree style: 

repli = flip $ concatMap . replicate
alternatively, without using the
replicate
function: 
repli :: [a] -> Int -> [a]
repli xs n = concatMap (take n . repeat) xs

or, using the list monad: 

repli :: [a] -> Int -> [a]
repli xs n = xs >>= replicate n
or, a more verbose solution without the use of
replicate
: 
repli :: [a] -> Int -> [a]
repli xs n = foldl (\acc e -> acc ++ repli' e n) [] xs
    where
      repli' _ 0 = []
      repli' x n = x : repli' x (n-1)

or, a version that does not use list concatenation: 

repli :: [a] -> Int -> [a]
repli [] _ = []
repli (x:xs) n = foldr (const (x:)) (repli xs n) [1..n]
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