The problem Reverse a List encourages to learn List.rev.

You are given a list of elements. Reverse the list without using the reverse function.

The simplest implementation is not very efficient one but it works good enough to pass the tests:

let rev (lst : List<'T>) =
let length = List.length lst
seq { for i in (length-1) .. -1 .. 0 do yield lst.[i] }
|> Seq.toList


The main method:

[<EntryPoint>]
let main argv =
readItemsFromInput ()
|> Seq.toList
|> rev
|> Seq.iter (fun v -> printfn "%A" v)
0 // return an integer exit code


Complexity of the solution is O(N²) because getting list item is O(N) and it should be done for very item of the list. The alternative can be using structure with O(1). You can try to do it as an exercise.