99 questions/Solutions/15: Difference between revisions
< 99 questions | Solutions
(discovered an interesting solution that was not on here) |
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or, a | or, a version that does not use list concatenation: | ||
<haskell> | <haskell> | ||
repli :: [a] -> Int -> [a] | repli :: [a] -> Int -> [a] | ||
repli [] _ = [] | repli [] _ = [] | ||
repli (x:xs) n = | repli (x:xs) n = foldr (const (x:)) (repli xs n) [1..n] | ||
</haskell> | </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) xsor, in Pointfree style:
repli = flip $ concatMap . replicatealternatively, without using the replicate function:
repli :: [a] -> Int -> [a]
repli xs n = concatMap (take n . repeat) xsor, using the list monad:
repli :: [a] -> Int -> [a]
repli xs n = xs >>= replicate nor, 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]