Difference between revisions of "99 questions/Solutions/15"
< 99 questions | Solutions
Jump to navigation
Jump to search
(6 intermediate revisions by 5 users not shown) | |||
Line 6: | Line 6: | ||
</haskell> |
</haskell> |
||
− | or, in Pointfree style: |
+ | or, in [[Pointfree]] style: |
<haskell> |
<haskell> |
||
repli = flip $ concatMap . replicate |
repli = flip $ concatMap . replicate |
||
</haskell> |
</haskell> |
||
+ | |||
+ | alternatively, without using the <hask>replicate</hask> function: |
||
+ | <haskell> |
||
+ | repli :: [a] -> Int -> [a] |
||
+ | repli xs n = concatMap (take n . repeat) xs |
||
+ | </haskell> |
||
+ | |||
+ | or, using the list monad: |
||
+ | <haskell> |
||
+ | repli :: [a] -> Int -> [a] |
||
+ | repli xs n = xs >>= replicate n |
||
+ | </haskell> |
||
+ | |||
+ | or, a more verbose solution without the use of <hask>replicate</hask>: |
||
+ | <haskell> |
||
+ | 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) |
||
+ | </haskell> |
||
+ | |||
+ | or, a version that does not use list concatenation: |
||
+ | <haskell> |
||
+ | repli :: [a] -> Int -> [a] |
||
+ | repli [] _ = [] |
||
+ | repli (x:xs) n = foldr (const (x:)) (repli xs n) [1..n] |
||
+ | </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) 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]