Difference between revisions of "99 questions/Solutions/15"

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(discovered an interesting solution that was not on here)
 
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or, a convoluted recursive solution that only uses cons:
 
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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>
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[[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]