Elixir从列表中删除重复项

Question

有人愿意使用Functional Programming和Elixir Constructs为List(X)中的重复值提供一些替代解决方案吗？

X = [1,26,3,40,5,6,6,7] # the 6 being the duplicate

我想到的用于解决此问题的库存解决方案是迭代列表(X),并添加到新的列表(Y),其中密钥尚不存在.

谢谢

Answer 1

Enum.uniq 做你想要的,例如:

iex(6)> Enum.uniq([1,26,3,40,5,6,6,7])
[1, 26, 3, 40, 5, 6, 7]

就你如何实现它而言,你可以编写一个递归函数,如下所示:

defmodule Test do
  def uniq(list) do
    uniq(list, HashSet.new)
  end

  defp uniq([x | rest], found) do
    if HashSet.member?(found, x) do
      uniq(rest, found)
    else
      [x | uniq(rest, HashSet.put(found, x))]
    end
  end

  defp uniq([], _) do
    []
  end
end

iex(3)> Test.uniq([1, 1, 2, 3, 4, 4, 1])
[1, 2, 3, 4]