# 99 questions/Solutions/15

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< 99 questions | Solutions(Difference between revisions)

(Added an alternative and more verbose solution) |
(discovered an interesting solution that was not on here) |
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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

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)

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