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

Question

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?

Answer 1

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.

Answer 2

It does not seem to improve anything but:

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