99 questions/Solutions/15: Difference between revisions

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repli xs n = concatMap (take n . repeat) xs
repli xs n = concatMap (take n . repeat) xs
</haskell>
</haskell>
or, using the list monad:
<haskell>
repli :: [a] -> Int -> [a]
repli xs n = xs >>= replicate n
</haskell>
or, a more verbose solution without the use of <hask>replicate</hask>:
<haskell>
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)
</haskell>
or, a version that does not use list concatenation:
<haskell>
repli :: [a] -> Int -> [a]
repli [] _ = []
repli (x:xs) n = foldr (const (x:)) (repli xs n) [1..n]
</haskell>
[[Category:Programming exercise spoilers]]

Latest revision as of 19:33, 18 January 2014

(**) 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]