链接列表分区功能和扭转业绩 [英] Linked list partition function and reversed results

问题描述

我写了这个F＃功能分区列表达到一定点并没有进一步的 - 就像在 takeWhile 交叉和分区。

I wrote this F# function to partition a list up to a certain point and no further -- much like a cross between takeWhile and partition.

let partitionWhile c l =
    let rec aux accl accr =
        match accr with
        | [] -> (accl, [])
        | h::t ->
            if c h then
                aux (h::accl) t
            else
                (accl, accr)
    aux [] l

唯一的问题是，采取项目是相反的：

The only problem is that the "taken" items are reversed:

> partitionWhile ((>=) 5) [1..10];;
val it : int list * int list = ([5; 4; 3; 2; 1], [6; 7; 8; 9; 10])

除了诉诸调用转，有没有办法这个功能可以写，将有第一个列表是正确的顺序？

Other than resorting to calling rev, is there a way this function could be written that would have the first list be in the correct order?

推荐答案

下面是一个基于延续的版本。这是尾递归并返回列表中的原始顺序。

Here's a continuation-based version. It's tail-recursive and returns the list in the original order.

let partitionWhileCps c l =
  let rec aux f = function
    | h::t when c h -> aux (fun (acc, l) -> f ((h::acc), l)) t
    | l -> f ([], l)
  aux id l

下面是一些基准一起去讨论以下布莱恩的回答（和蓄能​​器版本以供参考）：

Here are some benchmarks to go along with the discussion following Brian's answer (and the accumulator version for reference):

let partitionWhileAcc c l =
  let rec aux acc = function
    | h::t when c h -> aux (h::acc) t
    | l -> (List.rev acc, l)
  aux [] l

let test = 
  let l = List.init 10000000 id
  (fun f ->
    let r = f ((>) 9999999) l
    printfn "%A" r)

test partitionWhileCps // Real: 00:00:06.912, CPU: 00:00:07.347, GC gen0: 78, gen1: 65, gen2: 1
test partitionWhileAcc // Real: 00:00:03.755, CPU: 00:00:03.790, GC gen0: 52, gen1: 50, gen2: 1

厘泊平均〜7秒，加<​​/ code>〜4秒。总之，延续你买什么了这项工作。

Cps averaged ~7s, Acc ~4s. In short, continuations buy you nothing for this exercise.

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