Brush some notes of LeetCode with elixir

dynamic programming

Recently, I have been swiping dynamic programming questions. It just so happens that leetcode China has a "learning plan" function, which assigns me the task of several questions every day.

Summarized a set of small templates for dynamic programming. Take the question "Jumping Game" as an example

defmodule Solution do
  @spec can_jump(nums :: [integer]) :: boolean
  def can_jump([x]), do: true
  def can_jump([h | _] = nums) do

    # 首先构造初始状态，由于 List 的访问时间是 O(n) 的，我们先将其转换为 Map
    state = %{0 => h, nums: nums |> Enum.with_index() |> Enum.into(%{}, fn {v, k} -> {k, v} end)}

    # 用一个 Agent 来保存状态，因为 elixir 里面一般不使用全局变量
    Agent.start(fn -> state end, name: __MODULE__)

    r = dp(length(nums) - 1)

    # 结束后需要停止 Agent，否则状态会保存到下一个测试
    Agent.stop(__MODULE__)
    !!r
  end

  # dp 主逻辑
  defp dp(i) do
    find_and_store(i, fn %{nums: nums} -> nums[i] end, fn step -> 
      case dp(i - 1) do
        false -> false
        j ->
          if j >= i do
            max(j, i + step)
          else
            false
          end
      end
    end)
  end

  # 通用函数，计算并存储最新状态。fget 函数是在 Agent 里执行，再返回结果。
  # fdp 是状态转换的函数，其输入是 fget 的结果。
  defp find_and_store(i, fget, fdp) do
    case Agent.get(__MODULE__, fn m -> m[i] end) do
      nil -> 
        x = Agent.get(__MODULE__, fget) |> fdp.()
        Agent.update(__MODULE__, fn m -> Map.put(m, i, x) end)
        x

      x ->
        x
    end
  end
end

Everyone may say, you are recursive like this, won't you burst the stack? In fact, I am also trembling, for fear that not using tail recursion will cause the stack to explode, but I have not encountered such a situation yet. Maybe the optimization of beam is very good.