99 questions/Solutions/9
From HaskellWiki
(Difference between revisions)
(added one more implementation) |
|||
| Line 21: | Line 21: | ||
| otherwise = ([], (y:ys)) | | otherwise = ([], (y:ys)) | ||
(first,rest) = getReps xs | (first,rest) = getReps xs | ||
| + | </haskell> | ||
| + | |||
| + | Similarly, using <hask>splitAt</hask> and <hask>findIndex</hask>: | ||
| + | |||
| + | <haskell> | ||
| + | pack :: Eq a => [a] -> [[a]] | ||
| + | pack [] = [] | ||
| + | pack (x:xs) = (x:reps) : (pack rest) | ||
| + | where | ||
| + | (reps, rest) = maybe ([],xs) (\i -> splitAt i xs) (findIndex (/=x) xs) | ||
</haskell> | </haskell> | ||
Revision as of 16:55, 15 September 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
splitAt
findIndex
pack :: Eq a => [a] -> [[a]] pack [] = [] pack (x:xs) = (x:reps) : (pack rest) where (reps, rest) = maybe ([],xs) (\i -> splitAt i xs) (findIndex (/=x) xs)
group
Data.List
takeWhile
dropWhile
pack :: (Eq a) => [a] -> [[a]] pack [] = [] pack (x:xs) = (x : takeWhile (==x) xs) : pack (dropWhile (==x) xs)
