99 questions/Solutions/9
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(added one more implementation) 

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Revision as of 13:33, 24 June 2012
(**) 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) 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)