所有选择F#中的列表 - 更优雅和简单

Question

有人可以提出更好和/或更优雅的实现:

let each xs = 
    let rec each' acc left right = 
        match right with
        | [] -> acc
        | right ->  let new_left  = left @ [List.hd right]
                    let next   = List.tl right
                    let result = (List.hd right), left @ next
                    each' (result::acc) new_left next
    each' [] [] xs

它做到了:

> each [1..3];;
val it : (int * int list) list = [(3, [1; 2]); (2, [1; 3]); (1, [2; 3])]

此函数也可以反向返回结果.我们的想法是将所有元素作为元组,包含元素和rest元素列表.

Answer 1

这里的语义略有不同,但从你给出的例子Set可能是一个很好的选择:

let each xs =
    let X = set xs                           
    [ for x in xs -> (x, X - set [x]) ]


> fsi.AddPrinter( fun (x:Set<int>) -> sprintf "%A" (Set.to_list x))
> each [1..3];;
> val it : (int * Set<int>) list = [(1, [2; 3]); (2, [1; 3]); (3, [1; 2])]

// Edited as per comments.