Personal tools

99 questions/Solutions/9

From HaskellWiki

< 99 questions | Solutions(Difference between revisions)
Jump to: navigation, search
Line 23: Line 23:
 
                   | 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>:
Line 34: Line 34:
 
         (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>:
Line 88: Line 88:
 
| otherwise    = impl xs ([x]:p)
 
| otherwise    = impl xs ([x]:p)
 
</haskell>
 
</haskell>
 +
 +
[[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)