将项目列表拆分为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

我想做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.

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.

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

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

我写到这里为止，我不知道如何进行.

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 [] []

推荐答案

不发生堆栈溢出的实现:

Implementation which does not stack-overflows:

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

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