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]