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(added one more implementation)
m (oops mixed up my edge case)
Line 30: Line 30:
pack (x:xs) = (x:reps) : (pack rest)
pack (x:xs) = (x:reps) : (pack rest)
where
where
-
(reps, rest) = maybe ([],xs) (\i -> splitAt i xs) (findIndex (/=x) xs)
+
(reps, rest) = maybe (xs,[]) (\i -> splitAt i xs) (findIndex (/=x) xs)
</haskell>
</haskell>

Revision as of 17:06, 15 September 2010

(**) 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 [] = []

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)
This is implemented as
group
in
Data.List
. Another solution using
takeWhile
and
dropWhile
: 
pack :: (Eq a) => [a] -> [[a]]
pack [] = []
pack (x:xs) = (x : takeWhile (==x) xs) : pack (dropWhile (==x) xs)
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