Ruby中堆栈的未定义方法“大小”

Question

我正在尝试创建一个最小堆栈函数，但我想我不明白循环的大小，这是我的代码：

class Node
  attr_accessor :value, :next_node

  def initialize(value, next_node = nil)
    @value = value
    @next_node = next_node
  end
end
class LinkedList
  def initialize
    @head = nil
    @tail = nil
  end
  def add(number)
    new_node = Node.new(number)
    if @head.nil?
      @head = new_node
      @tail = new_node
    else
      @tail.next_node = new_node
      @tail = new_node
    end
  end
  def add_at(index, number)
    new_node = Node.new(number)
    if @head.nil?
      @head = new_node
      @tail = new_node
    else
      current = @head
      if index.zero?
        new_node.next_node = @head
        @head = new_node
        puts @head
      else
        (index - 1).times do
          current = current.next_node
        end 
        new_node.next_node = current.next_node
        current.next_node = new_node
        puts new_node
      end
    end
  end
  def get(index)
    current = @head
    index.times do
      current = current.next_node
    end
    current.value
  end
  def remove(index)
    if @head.nil?
      puts 'The list is already empty ^^'
    else
      if index.zero?
        @head = @head.next_node
        puts @head
      else
        current_at = @head
        (index - 1).times do
          current_at = current_at.next_node
        end
        tmp = current_at.next_node
        current_at.next_node = tmp.next_node
        puts tmp
      end
    end
  end
  def get_head
    @head
  end
end
class Stack
  def initialize
    @stack = LinkedList.new
  end
  def push(value)
    @stack.add_at(0, value)
  end
  def pop
    raise 'Stack is already empty' unless @stack
    temporal = @stack.get(0)
    @stack.remove(0)
    #temporal
  end

  def min
    current = @head
    (@stack.size).times do
      if current.next_node < current
        current = current.next_node
      end
    end
    current.value
  end

  def get_head
    @stack.get_head
  end
  def get_stack
    @stack
  end
end

stack = Stack.new
stack.push(3)
stack.push(5)
puts stack.min
# => 3

stack.pop
stack.push(7)
puts stack.min
# => 3

stack.push(2)
puts stack.min
# => 2

stack.pop
puts stack.min
# => 3

当然，这会抛出：

#<Node:0x0000560189f7b490>
undefined method `size' for #<LinkedList:0x0000560189f7b4e0>
(repl):90:in `min'
(repl):109:in `<main>'

使用https://repl.it编译

ruby 2.5.5p157（2019-03-15 修订版 67260）[x86_64-linux]

这样做的目的是创建一个堆栈，其中包含一个 min 方法，该方法返回堆栈中的最小数字。

我正在尝试使用 Do 循环创建 min 函数，但它不起作用，我做错了什么？

Answer 1

我做错了什么？

您正在通过@stack 呼叫size。但是，@stack 是一个 LinkedList，它没有 size 方法。