通过 Linked Stack 反转单链表答案

【问题标题】：Reverse a Singly linked list through Linked Stack通过 Linked Stack 反转单链表
【发布时间】：2015-12-30 05:36:10
【问题描述】：

是的，这是作业。我必须反转一个单链表。导师/朋友告诉我我应该使用堆栈，所以第一个进是最后一个出。我有这个，一点也不多，我不知道现在该做什么。 GameEntry 列表已提供。有没有朝着正确的方向推进？

    package project;

import java.util.LinkedList;
import dsaj.arrays.GameEntry;
import net.datastructures.*;

public class reverse {

public static void main(String[] args) 
{SinglyLinkedList<GameEntry>list = new SinglyLinkedList<GameEntry>();


list.addFirst(new GameEntry ("A", 0 ));
list.addFirst(new GameEntry ("B", 1 ));
list.addFirst(new GameEntry ("C", 2 ));
list.addFirst(new GameEntry ("D", 3 ));
list.addFirst(new GameEntry ("E", 4 ));
list.addFirst(new GameEntry ("F", 5 ));
list.addFirst(new GameEntry ("G", 6 ));
list.addFirst(new GameEntry ("H", 7 ));
list.addFirst(new GameEntry ("I", 8 ));
list.addFirst(new GameEntry ("J", 9 ));
System.out.printf("\nList before reverse\n %s", list);

SinglyLinkedList<GameEntry>list2 = new SinglyLinkedList<GameEntry>();
LinkedStack<GameEntry>stack =new LinkedStack<GameEntry>();                 
} 
  }

【问题讨论】：

一个足够成功的应该是堆栈数据结构以与添加（推送）方式相反的顺序删除（弹出）元素......Java中的LinkedList具有行为所需的结构和方法一个堆栈

标签： java stack reverse singly-linked-list

【解决方案1】：

迭代list，将值（在迭代时）推送到stack。然后将值（来自stack）弹出到list2，直到你没有来自stack 的项目。它可能看起来像

for (GameEntry ge : list) {
    stack.push(ge);
}
while (!stack.isEmpty()) {
    list2.add(stack.pop());
}

【讨论】：

【解决方案2】：

这两种方法应该足够了，你都必须在 SLL 类中

code here

     /**
     * Adds an element to the end of the list.
     * @param e  the new element to add
     */
    public void addLast(E e) {   // adds element e to the end 
      Node<E> newest = new Node<>(e, null);   
      if (isEmpty())
        head = newest;         // special case: previously empty list
      else
        tail.setNext(newest);        // new node after   existing tail
      tail = newest;                    // new node becomes the tail
      size++;
    }

    /**
     * Removes and returns the first element of the list.
     * @return the removed element (or null if empty)
     */
    public E removeFirst() {                   // removes and returns the first element
      if (isEmpty()) return null;              // nothing to remove
      E answer = head.getElement();
      head = head.getNext();                   // will become null if list had only one node
      size--;
      if (size == 0)
        tail = null;                           // special case as list is now empty
      return answer;
    }

【讨论】：