最小/最大和列表中最频繁的元素 [英] Min/Max and most frequent element of a list

问题描述

我必须编写一个程序来输出一个元组，该元组具有:非空列表的min和max以及最常出现的值. 特别是:

I have to write a program which give in output a tuple with: min and max of a not-empty list and the value that appears most often. In particular:

min_max [1;0;-1;2;0;-4] ==> (-4; 2)
min_max: int list -> (int * int)

mode [-1;2;1;2;5;-1;5;5;2] ==> 2
mode: int list -> int

这是我为max编写的代码(min几乎相等)，但是我该怎么做才能接收带有两个值的元组作为输出?

This is the code that I wrote for max (min is almost equal) but how can I do to receive as output a tuple with the two values?

let rec max_list xs =
    match xs with
    | [] -> failwith "xs" "Empty list"
    | [x] -> x
    | x1::x2::xs' -> max_list((max2 x1 x2)::xs');;

推荐答案

我将从

I'll take the first suggestion from @Mark Seemann's answer and run with it, in order to make it generic, working with any collection type, and handle the case of the empty collection sensibly.

let tryMinMax xs =
    Seq.fold (function
    | Some(mn, mx) -> fun i -> Some(min mn i, max mx i)
    | None         -> fun i -> Some(i, i) ) None xs

[1;0;-1;2;0;-4]
|> tryMinMax
// val it : (int * int) option = Some (-4, 2)

对于问题的最常见部分:

let mostFrequent xs =
    xs 
    |> Seq.countBy id 
    |> Seq.maxBy snd
    |> fst

[1;0;-1;2;0;-4]
|> mostFrequent
// val it : int = 0

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