Difference between revisions of "99 questions/Solutions/15"
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(added solution using list monad) |
m (Improved my version a little) |
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repli :: [a] -> Int -> [a] |
repli :: [a] -> Int -> [a] |
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repli xs n = xs >>= replicate n |
repli xs n = xs >>= replicate n |
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+ | </haskell> |
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+ | |||
+ | or, a more verbose solution without the use of <hask>replicate</hask>: |
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+ | <haskell> |
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+ | repli :: [a] -> Int -> [a] |
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+ | repli xs n = foldl (\acc e -> acc ++ repli' e n) [] xs |
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+ | where |
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+ | repli' _ 0 = [] |
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+ | repli' x n = x : repli' x (n-1) |
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+ | </haskell> |
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+ | |||
+ | or, a version that does not use list concatenation: |
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+ | <haskell> |
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+ | repli :: [a] -> Int -> [a] |
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+ | repli [] _ = [] |
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+ | repli (x:xs) n = foldr (const (x:)) (repli xs n) [1..n] |
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</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
alternatively, without using the replicate
function:
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
or, 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]