is it possible to do quicksort of a list with only one passing?
I am learning haskell and the function definition I see is:
quickSort (x : xs) = (quickSort less) ++ (x : equal) ++ (quickSort more)
where less = filter (< x) xs
equal = filter (== x) xs
more = filter (> x) xs
Is it possible to write it with only开发者_开发知识库 one traversal of the list, instead of 3?
You mean something like this?
quicksort [] = []
quicksort (x:xs) = quicksort less ++ (x : equal) ++ quicksort more
where (less, equal, more) = partition3 x xs
partition3 _ [] = ([], [], [])
partition3 x (y:ys) =
case compare y x of
LT -> (y:less, equal, more)
EQ -> (less, y:equal, more)
GT -> (less, equal, y:more)
where (less, equal, more) = partition3 x ys
Note that this isn't really quicksort, as the real quicksort is in-place.
It does not seem to improve anything but:
qs (x:xs) = let (a,b) = partition (< x) xs in (qs a) ++ [x] ++ (qs b)
Although late, here's a version that's supposed to not leak space as much (and seems to run about twice faster than the other 3-way version here):
qsort3 xs = go xs []
where
go (x:xs) zs = part x xs zs [] [] []
go [] zs = zs
part x [] zs a b c = go a ((x : b) ++ go c zs)
part x (y:ys) zs a b c =
case compare y x of
LT -> part x ys zs (y:a) b c
EQ -> part x ys zs a (y:b) c
GT -> part x ys zs a b (y:c)
This addresses the possible problem with using tuples, where let (a,b) = ...
is actually translated into let t= ...; a=fst t; b=snd t
which leads to the situation where even after a
has been consumed and processed, it is still kept around alive, as part of the tuple t
, for b
to be read from it - though of course completely unnecessary. This is known as "Wadler pair space leak" problem. Or maybe GHC (with -O2
) is smarter than that. :)
Also this apparently uses difference lists approach (thanks, hammar) which also makes it a bit more efficient (about twice faster than the version using tuples). I think part
uses accumulator parameters, as it builds them in reversed order.
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