99 questions/Solutions/9
From HaskellWiki
(Difference between revisions)
| Line 1: | Line 1: | ||
(**) Pack consecutive duplicates of list elements into sublists. | (**) Pack consecutive duplicates of list elements into sublists. | ||
| + | |||
If a list contains repeated elements they should be placed in separate sublists. | If a list contains repeated elements they should be placed in separate sublists. | ||
| Line 9: | Line 10: | ||
A more verbose solution is | A more verbose solution is | ||
| + | |||
<haskell> | <haskell> | ||
pack :: Eq a => [a] -> [[a]] | pack :: Eq a => [a] -> [[a]] | ||
| Line 22: | Line 24: | ||
This is implemented as <hask>group</hask> in <hask>Data.List</hask>. | This is implemented as <hask>group</hask> in <hask>Data.List</hask>. | ||
| + | |||
| + | Another solution using <hask>takeWhile</hask> and <hask>dropWhile</hask>: | ||
| + | |||
| + | <haskell> | ||
| + | pack :: (Eq a) => [a] -> [[a]] | ||
| + | pack [] = [] | ||
| + | pack (x:xs) = (x : takeWhile (==x) xs) : pack (dropWhile (==x) xs) | ||
| + | </haskell> | ||
Revision as of 03:00, 14 July 2010
(**) Pack consecutive duplicates of list elements into sublists.
If a list contains repeated elements they should be placed in separate sublists.
pack (x:xs) = let (first,rest) = span (==x) xs in (x:first) : pack rest pack [] = []
A more verbose solution is
pack :: Eq a => [a] -> [[a]] pack [] = [] pack (x:xs) = (x:first) : pack rest where getReps [] = ([], []) getReps (y:ys) | y == x = let (f,r) = getReps ys in (y:f, r) | otherwise = ([], (y:ys)) (first,rest) = getReps xs
group
Data.List
takeWhile
dropWhile
pack :: (Eq a) => [a] -> [[a]] pack [] = [] pack (x:xs) = (x : takeWhile (==x) xs) : pack (dropWhile (==x) xs)
