# 99 questions/Solutions/15

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< 99 questions | Solutions(Difference between revisions)

(discovered an interesting solution that was not on here) |
m (Improved my version a little) |
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Line 32: | Line 32: | ||

</haskell> |
</haskell> |
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− | or, a convoluted recursive solution that only uses cons: |
+ | or, a version that does not use list concatenation: |

<haskell> |
<haskell> |
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repli :: [a] -> Int -> [a] |
repli :: [a] -> Int -> [a] |
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repli [] _ = [] |
repli [] _ = [] |
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− | 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> |

## Revision as of 16:55, 19 November 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 version that does not use list concatenation:

repli :: [a] -> Int -> [a] repli [] _ = [] repli (x:xs) n = foldr (const (x:)) (repli xs n) [1..n]