99 questions/Solutions/9: Difference between revisions
< 99 questions | Solutions
added one more implementation |
m oops mixed up my edge case |
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pack (x:xs) = (x:reps) : (pack rest) | pack (x:xs) = (x:reps) : (pack rest) | ||
where | where | ||
(reps, rest) = maybe ([] | (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)