是否可以只通过一次就对列表进行快速排序? [英] is it possible to do quicksort of a list with only one passing?

问题描述

我正在学习haskell，我看到的函数定义是:

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

是否可以只遍历列表一次而不是 3 次来编写它?

Is it possible to write it with only one traversal of the list, instead of 3?

推荐答案

虽然晚了，但这里的版本应该不会泄漏太多空间(并且似乎比其他 3 个版本快两倍-方式版本在这里):

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)

这解决了使用元组可能存在的问题，其中 let (a,b) = ... 实际上被翻译成 let t= ...;a=fst t;b=snd t 这导致即使在 a 被消费和处理后，它仍然保持活动状态，作为元组 t 的一部分, 用于从它读取 b - 尽管当然完全没有必要.这被称为Wadler 对空间泄漏"问题.或者也许 GHC(带有 -O2)比这更聪明.:)

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. :)

此外，这显然使用了差异列表方法(感谢，hammar)，这也使其效率更高(比使用元组的版本快大约两倍).我认为 part 使用累加器参数，因为它以相反的顺序构建它们.

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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