# 99 questions/Solutions/9

### From HaskellWiki

< 99 questions | Solutions(Difference between revisions)

Mikesongming (Talk | contribs) m |
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func x (y:xs) = |
func x (y:xs) = |
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if x == (head y) then ((x:y):xs) else ([x]:y:xs) |
if x == (head y) then ((x:y):xs) else ([x]:y:xs) |
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+ | </haskell> |
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+ | |||

+ | A simple solution: |
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+ | <haskell> |
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+ | pack :: (Eq a) => [a] -> [[a]] |
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+ | pack [] = [] |
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+ | pack [x] = [[x]] |
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+ | pack (x:xs) = if x `elem` (head (pack xs)) |
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+ | then (x:(head (pack xs))):(tail (pack xs)) |
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+ | else [x]:(pack xs) |
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</haskell> |
</haskell> |

## Revision as of 16:16, 18 April 2011

(**) 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 [] = []

group

Data.List

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)

takeWhile

dropWhile

pack :: (Eq a) => [a] -> [[a]] pack [] = [] pack (x:xs) = (x : takeWhile (==x) xs) : pack (dropWhile (==x) xs)

foldr

pack :: (Eq a) => [a] -> [[a]] pack = foldr func [] where func x [] = [[x]] func x (y:xs) = if x == (head y) then ((x:y):xs) else ([x]:y:xs)

A simple solution:

pack :: (Eq a) => [a] -> [[a]] pack [] = [] pack [x] = [[x]] pack (x:xs) = if x `elem` (head (pack xs)) then (x:(head (pack xs))):(tail (pack xs)) else [x]:(pack xs)