Difference between revisions of "99 questions/Solutions/9"
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(added one more implementation) |
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| otherwise = ([], (y:ys)) |
| otherwise = ([], (y:ys)) |
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(first,rest) = getReps xs |
(first,rest) = getReps xs |
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+ | </haskell> |
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+ | |||
+ | Similarly, using <hask>splitAt</hask> and <hask>findIndex</hask>: |
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+ | |||
+ | <haskell> |
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+ | pack :: Eq a => [a] -> [[a]] |
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+ | pack [] = [] |
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+ | pack (x:xs) = (x:reps) : (pack rest) |
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+ | where |
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+ | (reps, rest) = maybe ([],xs) (\i -> splitAt i xs) (findIndex (/=x) xs) |
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</haskell> |
</haskell> |
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Revision as of 16:55, 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)