不使用数组方法反转 LinkedList [英] Reverse a LinkedList WITHOUT using array methods

问题描述

所以，我正在处理一个任务，其中 LinkedListNode 类生成一个链表并将值打印给用户.我应该做的是创建一个 Stack 类来推送"、弹出"和反转链表.问题是，我们无法使用 .push、.pop 或其他任何东西的数组方法.

So, I'm working on an assignment where a LinkedListNode class generates a linked list and prints the values to the user. What I am supposed to do is create a Stack class to 'push', 'pop', and reverse the linked list. The thing is, we are unable to use the array methods of .push, .pop, or anything else.

一天多来，我一直在思考并试图查找要做什么，但不知道从哪里开始.有人可以请我指出正确的方向，让我开始使用推送方法，这样我就可以得到一些牵引力，也许能赶上?

I have been thinking and trying to look up what to do for over a day and have no clue on where to even start. Could someone please point me in the right direction of just getting the push method started so I can get some traction and maybe catch on?

给出的 LinkedListNode 类:

LinkedListNode class given:

class LinkedListNode
  attr_accessor :value, :next_node

  def initialize(value, next_node=nil)
      @value = value
      @next_node = next_node
  end
end

给出的 Stack 类的起始骨架:

Starting skeleton of Stack class given:

class Stack
  attr_reader :data

  def initialize
      @data = nil
  end

  # Push an item onto the stack
  def push(element)
  # IMPLEMENT ME!
  end

  # Pop an item off the stack.  
  # Remove the last item that was pushed onto the
  # stack and return it to the user
  def pop
      # IMPLEMENT ME
  end
end

我不是想直接询问答案，我只是不知道该去哪里.在此先感谢您的帮助！

I am not trying to ask for the answer straight out, I just can't figure out where to go. Thank you in advance for any help!

推荐答案

您可以使用 Linked List 数据类型而不是数组来实现 Stack.我相信这样的事情会奏效:

You can use the Linked List datatype, rather than an array, to implement a Stack. I believe something like this would work:

def push(element)
  if data.nil?
    data = LinkedListNode.new(element, nil)
  else
    data = LinkedListNode.new(element, data)
  end
end

def pop
  # grab the top piece of data
  popped = data.value
  # shift the data
  data = data.next_node
end

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