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99 questions/Solutions/15

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(Added an alternative and more verbose solution)
(discovered an interesting solution that was not on here)
Line 30: Line 30:
 
repli' _ 0 = []
 
repli' _ 0 = []
 
repli' x n = x : repli' x (n-1)
 
repli' x n = x : repli' x (n-1)
  +
</haskell>
  +
  +
or, a convoluted recursive solution that only uses cons:
  +
<haskell>
  +
repli :: [a] -> Int -> [a]
  +
repli [] _ = []
  +
repli (x:xs) n = foldl (\f _ -> (x:) . f) (repli xs) [1..n] n
 
</haskell>
 
</haskell>

Revision as of 09:42, 23 January 2012

(**) 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 convoluted recursive solution that only uses cons:

repli :: [a] -> Int -> [a]
repli [] _ = []
repli (x:xs) n = foldl (\f _ -> (x:) . f) (repli xs) [1..n] n