I'm working on Ocaml.org 99 problems, and I solved the run-length decode one.
Here is the solution given by the site:
let decode l = let rec many acc n x = if n = 0 then acc else many (x :: acc) (n-1) x in let rec aux acc = function | [] -> acc | One x :: t -> aux (x :: acc) t | Many (n,x) :: t -> aux (many acc n x) t in aux [] (List.rev l);;
And here is mine:
let decode l =
let rec aux acc = function
| [] -> acc
| One elem :: t -> aux (elem :: acc) t
| Many(cnt, elem) :: t -> if cnt > 0 then aux (elem::acc) (Many(cnt - 1, elem) :: t) else aux acc t in
List.rev (aux [] l);;
Test:
type 'a rld =
| One of 'a
| Many of int * 'a;;
(* result expected : ["a"; "a"; "a"; "a"; "b"; "c"; "c"; "a"; "a"; "d"; "e"; "e"; "e"; "e"] *)
decode [Many (4,"a"); One "b"; Many (2,"c"); Many (2,"a"); One "d"; Many (4,"e")];;
Could my solution become more optimal?
