99 questions/Solutions/15: Difference between revisions
< 99 questions | Solutions
Added an alternative and more verbose solution |
discovered an interesting solution that was not on here |
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| Line 30: | Line 30: | ||
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) 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 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