Ocaml双向链表：从双向链表中删除满足条件的节点

Question

我们有一个双向链接列表，定义为：

type 'a llist =
  | Nil
  | Cons of (float *'a) * 'a lcell * 'a lcell
and 'a lcell = ('a llist) ref

我已经实现了add head功能：

let add_head x head = 
  match !(!head) with
  | Nil -> head := !(singleton x)
  | Cons (e, previous, next) -> 
      let temp = Cons (x, ref Nil, !head) in
      previous := temp;
      head := previous;; 

注意，要实现添加头，我使用了单例功能

let singleton (init: float * 'a): 'a lcell ref =
  let l = ref (Cons (init, ref Nil, ref Nil)) in
  let front = ref l in
  front

我的问题是，当我尝试删除一个元素时，我试图编写一个删除函数remove: (float -> bool) -> 'a lcell ref -> unit，以便remove p head删除其时间戳满足谓词p: float -> bool的第一个节点。如果没有节点的时间戳满足该谓词，则列表应保持不变。

这是我到目前为止所拥有的：

let remove p head =
  let rec remove' ll =
    match !ll with 
    | Nil -> head := !head
    | Cons ( (d,_), previous, next) ->
        if p d then
          match (!previous, !next) with 
          | (Nil, Nil) -> head := ref Nil   (* empty list*)
          | (Nil, Cons ( d1, p1, n1)) -> (* this is the head, remove it and reassign head*)
              head := next; 
              p1 := Nil
          | (Cons ( d2, p2, n2), Cons ( d1, p1, n1)) -> (* this is middle, remove it and fix pointers of previous and next*)
              n2 := !next;
              p1 := !previous 
          | (Cons ( d1, p1, n1), Nil) -> (* this is tail, remove it and make previous one the tail*)
              n1:= Nil 
        else remove' next              
  in
  remove' !head

我在删除列表中间的项目时遇到麻烦，即不是头部或尾部。我也无法删除多个元素。有人可以帮助我吗，我认为我的比赛案件中缺少一些东西。

Answer 1

[当您在比赛陈述中做出负面决定时，您就一团糟您必须替换上一个和下一个，而不是n2和p1应该是

| Cons(d2, p2, n2), Cons (d1, p1, n1) ->
`previous := Cons(d2, p2, next);`
`next := Cons(d1, previous, n1);