# 99 questions/Solutions/9

### From HaskellWiki

(**) 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

Similarly, using

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)

Another solution using

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) pack' [] = [] pack' [x] = [[x]] pack' (x:xs) | x == head h_p_xs = (x:h_p_xs):t_p_hs | otherwise = [x]:p_xs where p_xs@(h_p_xs:t_p_hs) = pack' xs

myPack [] = [] myPack (y:ys) = impl ys [[y]] where impl [] packed = packed impl (x:xs) packed | x == (head (last packed)) = impl xs ((init packed) ++ [x:(last packed)]) | otherwise = impl xs (packed ++ [[x]]) myPack' [] = [] myPack' (y:ys) = reverse $ impl ys [[y]] where impl [] packed = packed impl (x:xs) p@(z:zs) | x == (head z) = impl xs ((x:z):zs) | otherwise = impl xs ([x]:p)