将项目列表拆分为包含奇数和偶数索引项目的两个列表 [英] Splitting a list of items into two lists of odd and even indexed items

问题描述

我想创建一个接受列表并返回两个列表的函数：第一个包含每个奇数项，第二个包含每个偶数项。

例如，给定 [1; 2; 4; 6; 7; 9] ，我想返回 [[1; 4; 7]; [2; 6; 9]] 。

我已经写了这篇文章，但我不知道如何进展。

  let splitList list = 
 let rec splitOdd oList list1 list2 = 
 match oList with 
 | []  - > [] 
 | head :: tail  - > splitEven tail（list1 :: head）list2 
和splitEven oList list1 list2 = 
 match oList with 
 | []  - > [] 
 | head :: tail  - > splitOdd tail list1（list2 :: head）
 splitOdd list [] [] 
  

解决

  let splitList list = List.foldBack（fun x （l，r） - > x :: r，l）list（[]，[]）
  

I would like to make a function that accepts a list and returns two lists: the first contains every odd item, and the second contains every even item.

For example, given [1;2;4;6;7;9], I would like to return [ [1;4;7] ; [2;6;9] ].

I have written this so far and I do not know how to progress.

let splitList list =
    let rec splitOdd oList list1 list2 =
        match oList with
        | [] -> []
        | head :: tail -> splitEven tail (list1::head) list2
    and splitEven oList list1 list2 =
        match oList with
        | [] -> []
        | head :: tail -> splitOdd tail list1 (list2::head)
    splitOdd list [] []

解决方案

Implementation which does not stack-overflows:

let splitList list = List.foldBack (fun x (l,r) -> x::r, l) list ([],[])

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