99 questions/Solutions/9

(**) 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 [] = []

This is implemented as group in 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 and 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 and dropWhile:

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

Or we can use foldr to implement this:

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)