Difference between revisions of "99 questions/Solutions/9"

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(Added a new solution using break)
(Undo revision 39658 by Tyr (Talk))
Line 61: Line 61:
 
then (x:(head (pack xs))):(tail (pack xs))
 
then (x:(head (pack xs))):(tail (pack xs))
 
else [x]:(pack xs)
 
else [x]:(pack xs)
</haskell>
 
 
 
A simpler solution which is similar to the takeWhile/dropWhile solution, but in one step.
 
<haskell>
 
pack :: (Eq a) => [a] -> [[a]]
 
pack (x:xs) = (x:xs') : (pack ys)
 
where (xs',ys) = break (/=x) xs
 
pack [] = []
 
 
</haskell>
 
</haskell>

Revision as of 14:41, 24 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 [] = []

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)

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)