Difference between revisions of "99 questions/Solutions/9"
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m (oops mixed up my edge case) |
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pack [] = [] |
pack [] = [] |
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</haskell> |
</haskell> |
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+ | |||
⚫ | |||
A more verbose solution is |
A more verbose solution is |
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(reps, rest) = maybe (xs,[]) (\i -> splitAt i xs) (findIndex (/=x) xs) |
(reps, rest) = maybe (xs,[]) (\i -> splitAt i xs) (findIndex (/=x) xs) |
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</haskell> |
</haskell> |
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− | |||
⚫ | |||
Another solution using <hask>takeWhile</hask> and <hask>dropWhile</hask>: |
Another solution using <hask>takeWhile</hask> and <hask>dropWhile</hask>: |
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Line 41: | Line 41: | ||
pack [] = [] |
pack [] = [] |
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pack (x:xs) = (x : takeWhile (==x) xs) : pack (dropWhile (==x) xs) |
pack (x:xs) = (x : takeWhile (==x) xs) : pack (dropWhile (==x) xs) |
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+ | </haskell> |
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+ | |||
+ | Or we can use <hask>foldr</hask> to implement this: |
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+ | |||
+ | <haskell> |
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+ | pack :: (Eq a) => [a] -> [[a]] |
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+ | pack = foldr func [] |
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+ | where func x [] = [[x]] |
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+ | func x (y:xs) = |
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+ | if x == (head y) then ((x:y):xs) else ([x]:y:xs) |
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</haskell> |
</haskell> |
Revision as of 10:55, 17 November 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 [] = []
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)