括号匹配算法

match.ex

问题：
"(){}[]"   =>  True
"([{}])"   =>  True
"(}"       =>  False
"[(])"     =>  False
"[({})](]" =>  False
基本思想：在算法中设置一个栈，每读入一个括号，若是右括号，则或者与栈顶匹配的左括号相互消解，或者是不合法的情况；若是左括号，则直接压入栈中。
若括号匹配，在算法的开始和结束时，栈都应该是空的。

defmodule Challenge do
  def valid_braces(braces), do: valid_braces(braces, [])

  def valid_braces("(" <> rest, opened), do: valid_braces(rest, ["(" | opened])
  def valid_braces("[" <> rest, opened), do: valid_braces(rest, ["[" | opened])
  def valid_braces("{" <> rest, opened), do: valid_braces(rest, ["{" | opened])

  def valid_braces(")" <> rest, ["(" | opened]), do: valid_braces(rest, opened)
  def valid_braces("]" <> rest, ["[" | opened]), do: valid_braces(rest, opened)
  def valid_braces("}" <> rest, ["{" | opened]), do: valid_braces(rest, opened)

  def valid_braces("", []), do: true
  def valid_braces(_, _), do: false
end

defmodule Challenge do
  def valid_braces(braces, length \\ 0) do
    length = String.length(braces)
    new_braces = braces
                |> String.replace("()", "")
                |> String.replace("{}", "")
                |> String.replace("[]", "")
    
    cond do
      (String.length(new_braces) == length) ->
        false
      (new_braces == "") ->
        true
      true ->
        valid_braces(new_braces, length)
    end
    
  end
end