99 questions/Solutions/15
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(Difference between revisions)
(added solution using list monad) |
(Added an alternative and more verbose solution) |
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| Line 21: | Line 21: | ||
repli :: [a] -> Int -> [a] | repli :: [a] -> Int -> [a] | ||
repli xs n = xs >>= replicate n | 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> | </haskell> | ||
Revision as of 05:11, 10 August 2011
(**) 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
replicate
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
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)
