Ocaml中的列表倒序 [英] List reversing in Ocaml

问题描述

如果我们假设元素从0开始计数，那么如何反转列表的子列表.我希望解决方案是手动编码"的.我在完成这项任务时遇到了很大的问题.

How to reverse even sublists of a list if we assume that we count elements from 0. I want the solution to be "manually-coded". I've got a big problem with this task.

例如:

Function([[1;2;3] ; [2;3] ; [1;2;3] ; [5;6;7]])

返回:

([[3;2;1] ; [2;3] ; [3;2;1] ; [5;6;7]])

我已经创建了一个可以反转单个列表的函数:

I already created a function that reverse a single list:

let rev =
  let rec rev_append acc l =
    match l with
      [] -> acc
    | h::t -> rev_append (h::acc) t in
  fun l -> rev_append [] l;;

但是现在我被困住了.

But now i am stuck.

推荐答案

let rev_list l =
  let rec rev_acc acc = function
    | [] -> acc
    | hd::tl -> rev_acc (hd::acc) tl
  in 
  rev_acc [] l

let rev_even l = 
  let rec rev i acc = function
    | [] -> rev_list acc
    | hd::tl ->
      if i mod 2 = 0 then rev (i+1) ((rev_list hd)::acc) tl
      else rev (i+1) (hd::acc) tl
  in 
  rev 0 [] l

请注意，它们都是尾递归的

note that they are all tail-recursive

修改

对诺兰的建议

尾递归在函数式编程和OCaml中非常重要.请记住.

tail-recursive is quite important in functional programming and OCaml. Please bear in mind.

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