99 questions/Solutions/15
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(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
