is it possible to do quicksort of a list with only one passing?

问题

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 one traversal of the list, instead of 3?

回答1:

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.

回答2:

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.

回答3:

It does not seem to improve anything but:

qs (x:xs) = let (a,b) = partition (< x) xs in (qs a) ++ [x] ++ (qs b)

来源：https://stackoverflow.com/questions/7641936/is-it-possible-to-do-quicksort-of-a-list-with-only-one-passing