在长生不老药中,如何一次处理两个相邻的列表项?

您可以通过常规链表递归并解构当前头和下一个头。要查找的特殊情况就在基本情况之前(即没有“下一个id”或类似id的项目)。

例如,如果我们有一个地图列表,比如:

  my_list = [
    %{id: 1, next: nil, previous: nil},
    %{id: 2, next: nil, previous: nil},
    ...
  ]


  def doubly_link([]), do: []

  def doubly_link([head | []]) do
    head = %{head | next: nil}

    [head | doubly_link([])]
  end

  def doubly_link([head | [next_head | tail]]) do
    head = %{head | next: next_head.id}
    next_head = %{next_head | previous: head.id}

    [head | doubly_link([next_head | tail])]
  end

然后打电话 doubly_link(my_list) 会产生:

[
  %{id: 1, previous: nil, next: 2},
  %{id: 2, previous: 1, next: 3},
  ....,
  %{id: n, previous: prior_to_n_id, next: nil}
]

另一种方法是 Enum.with_index/2

input = Enum.map(0..4, & %{id: &1})
head = -1
tail = Enum.count(input)

input
|> Enum.with_index(head)
|> Enum.with_index(head + 2)
|> Enum.map(fn
  {{value, ^head}, next} ->
    Map.put(value, :next, next)
  {{value, prev}, ^tail} ->
    Map.put(value, :prev, prev)
  {{value, prev}, next} ->
    value
    |> Map.put(:prev, prev)
    |> Map.put(:next, next)
end)

#⇒ [
#    %{id: 0, next: 1},
#    %{id: 1, next: 2, prev: 0},
#    %{id: 2, next: 3, prev: 1},
#    %{id: 3, next: 4, prev: 2},
#    %{id: 4, prev: 3}
# ]