Min/Max and most frequent element of a list

问题

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

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');;

回答1:

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)

For the most frequent part of the question:

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

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

回答2:

let minMax xs =
    xs
    |> List.fold
        (fun (mn, mx) i -> min mn i, max mx i)
        (System.Int32.MaxValue, System.Int32.MinValue)

Not particularly efficient, but fun to write:

let mode xs =
    xs
    |> List.groupBy id
    |> List.map (fun (i, is) -> i, Seq.length is)
    |> List.maxBy snd
    |> fst

回答3:

Option without the use of standard modules:

open System
let tryMinMax xs = 
    let rec minMax xs mn mx = 
        match xs with | [] -> mn, mx | h::t -> minMax t (min mn h) (max mx h)
    match xs with | [] -> None | _ -> Some(minMax xs Int32.MaxValue Int32.MinValue)

dotnetfiddle

On the second question - show their attempts to solve.

来源：https://stackoverflow.com/questions/33529942/min-max-and-most-frequent-element-of-a-list

标签