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

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| otherwise = ([], (y:ys))
 
| otherwise = ([], (y:ys))
 
(first,rest) = getReps xs
 
(first,rest) = getReps xs
</haskell>
+
</haskell><br>
   
 
Similarly, using <hask>splitAt</hask>&nbsp;and <hask>findIndex</hask>:
 
Similarly, using <hask>splitAt</hask>&nbsp;and <hask>findIndex</hask>:
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(reps, rest) = maybe (xs,[]) (\i -> splitAt i xs)
 
(reps, rest) = maybe (xs,[]) (\i -> splitAt i xs)
 
(findIndex (/=x) xs)
 
(findIndex (/=x) xs)
</haskell>
+
</haskell><br>
   
 
Another solution using <hask>takeWhile</hask>&nbsp;and <hask>dropWhile</hask>:
 
Another solution using <hask>takeWhile</hask>&nbsp;and <hask>dropWhile</hask>:
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| otherwise = impl xs ([x]:p)
 
| otherwise = impl xs ([x]:p)
 
</haskell>
 
</haskell>
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[[Category:Programming exercise spoilers]]

Revision as of 13:48, 15 February 2015

(**) 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)

pack' [] = []
pack' [x] = [[x]]
pack' (x:xs)
    | x == head  h_p_xs = (x:h_p_xs):t_p_hs
    | otherwise         = [x]:p_xs
    where p_xs@(h_p_xs:t_p_hs) = pack' xs
myPack [] = []
myPack (y:ys) = impl ys [[y]]
	where
		impl [] packed = packed
		impl (x:xs) packed 
			| x == (head (last packed)) = impl xs ((init packed) ++ [x:(last packed)])
			| otherwise		    = impl xs (packed ++ [[x]])

myPack' [] = []									
myPack' (y:ys) = reverse $ impl ys [[y]]
	where
		impl [] packed = packed
		impl (x:xs) p@(z:zs) 
			| x == (head z) = impl xs ((x:z):zs) 
			| otherwise     = impl xs ([x]:p)