為什么從我的LinkedList中刪除所有元素？

Question

為什么我的remove方法從我的雙向鏈接列表中刪除每個元素？ 如果我剔除if / else語句，則可以成功刪除中間元素，但列表開頭或結尾的元素仍然保留。 但是，我添加了if / else語句來處理頭部和尾部的元素，不幸的是，此方法現在刪除了列表中的每個元素。 我做錯了什么？

package week6;

import java.util.Iterator;

public class DblLinkedList<E>
{
   private LinkEntry<E> head = null;
   private LinkEntry<E> tail = null;
   private int size = 0;

   public DblLinkedList()
   {
      head = tail = null;
   }

   public boolean is_empty()
   {
      if (head == null) 
          return true;
      return false;
   }

   public int size()
   {
       int count = 0;
       for (LinkEntry<E> current = head; current != null; current = current.next)
           count++;
       return count;
   }

   public boolean add(E e)
   {   
      LinkEntry<E> new_element = new LinkEntry<E>();
      new_element.element = e;

          if (head == null)
          {
              new_element.next = head;
              head = new_element;
              tail = head;
          }
          else
          {
              tail.next = new_element;
              new_element.previous = tail;
              tail = new_element;
          }
          return true;
   }

   public void remove(int n)
   {
       LinkEntry<E> remove_this = new LinkEntry<E>();

       //if nothing comes before remove_this, set the head to equal the element after remove_this
       if (remove_this.previous == null)
           head = remove_this.next;

       //if nothing comes after remove_this, set the tail equal to the element before remove_this
       else if (remove_this.next == null)
           tail = remove_this.previous;
       //otherwise set the next element's previous pointer to the element before remove_this
       else
       {
           //if remove_this is located in the middle of the list, enter this loop until it is
           //found, then remove it, closing the gap afterwards.
           int i = 0;
           for (remove_this = head; remove_this != null; remove_this = remove_this.next)
           {
               //if i == n, stop and delete 'remove_this' from the list
               if (i == n)
               {               
                   //set the previous element's next to the element that comes after remove_this
                   remove_this.previous.next = remove_this.next;
                   //set the element after remove_this' previous pointer to the element before remove_this
                   remove_this.next.previous = remove_this.previous;
                   break;
               }
               //if i != n, keep iterating through the list
               i++; 
           }
       }
   }

   /*
    * Print the doubly linked list starting at the beginning.
    */
   public void print_from_beginning()
   {
      LinkEntry<E> current = new LinkEntry<E>();
      for (current = head; current != null; current = current.next)
      {
          System.out.print(current.element + " ");
      }
   }

   /*
    * Print the doubly linked list starting the end.
    */
   public void print_from_end()
   {
      LinkEntry<E> current = new LinkEntry<E>();
      for (current = tail; current != null; current = current.previous)
      {
          System.out.print(current.element + " ");
      }
   }

   /* ------------------------------------------------------------------- */
   /* Inner classes                                                      */
   protected class LinkEntry<E>
   {
      protected E element;
      protected LinkEntry<E> next;
      protected LinkEntry<E> previous;

      protected LinkEntry() { element = null; next = previous = null; }
   }
   /* ------------------------------------------------------------------- */
   protected class DblLinkedListImplIterate<E> implements Iterator<E>
   {

       protected LinkEntry<E> next;

       protected DblLinkedListImplIterate()
       {
           next = (LinkEntry<E>) head;
       }

    @Override
    public boolean hasNext() {
        // TODO Auto-generated method stub
        return false;
    }

    @Override
    public E next() {
        // TODO Auto-generated method stub
        return null;
    }

    @Override
    public void remove() {
        // TODO Auto-generated method stub
    }   
   }
}

我測試其方法的主類是：

package week6;

public class App {

    public static <E> void main(String[] args) {

        DblLinkedList<String> list = new DblLinkedList<String>();

        list.add("Bill");
        list.add("Rohan");
        list.add("James");
        list.add("Krishna");
        list.add("Javier");
        list.add("Lisa");

        System.out.println("List size after all names are added: " + list.size());

        //a. Print the linked list starting at the beginning.
        System.out.println("\nPrint the linked list starting at the beginning:");
        list.print_from_beginning();
        System.out.println();

        //b. Print the linked list starting at the end.
        System.out.println("\nPrint the linked list starting at the end:");
        list.print_from_end();
        System.out.println();

        //c. Remove Bill and print the linked list starting from beginning.
        System.out.println("\nRemove Bill and print the linked list starting from beginning:");
        list.remove(1);
        list.print_from_beginning();
        System.out.println();

        //d. Remove Lisa and print the linked list starting from end.
        System.out.println("\nRemove Lisa and print the linked list starting from end:");
        list.remove(5);
        list.print_from_end();
        System.out.println();

        //e. Remove Krishna and print the linked list starting from the beginning.
        System.out.println("\nRemove Krishna and print the linked list starting from the beginning:");
        list.remove(2);
        list.print_from_beginning();
        System.out.println();

        System.out.println("\nList size: " + list.size());

    }
}

運行程序后打印出的結果是：

List size after all names are added: 6

Print the linked list starting at the beginning:
Bill Rohan James Krishna Javier Lisa 

Print the linked list starting at the end:
Lisa Javier Krishna James Rohan Bill 

Remove Bill and print the linked list starting from beginning:


Remove Lisa and print the linked list starting from end:
Lisa Javier Krishna James Rohan Bill


Remove Krishna and print the linked list starting from the beginning:


List size: 0

Answer 1

通過添加兩個if塊來處理第一個和最后一個，除了之前的內容外，您還添加了兩個要運行的可執行塊。 現在，所有三個塊都一起執行。 我相信您正在嘗試執行以下操作（一次僅執行一個程序段）：

   public void remove(int n)
   {
       LinkEntry<E> remove_this = new LinkEntry<E>();
      //if remove_this is located in the middle of the list, enter this loop until it is
       //found, then remove it, closing the gap afterwards.
       int i = 0;
       boolean removed = false;
       remove_this = head;

       while(removed == false){
           //if nothing comes before remove_this, set the head to equal the element after remove_this
           if (remove_this.previous == null){
               head = remove_this.next;
               head.previous = null;
               removed = true;
           }

           //if nothing comes after remove_this, set the tail equal to the element before remove_this
           else if (remove_this.next == null){
               tail = remove_this.previous;
               tail.next = null;
               removed = true;
           }
           //otherwise set the next element's previous pointer to the element before remove_this
           else{
               //if i == n, stop and delete 'remove_this' from the list
               if (i == n) {               
                   //set the previous element's next to the element that comes after remove_this
                   remove_this.previous.next = remove_this.next;
                   //set the element after remove_this' previous pointer to the element before remove_this
                   remove_this.next.previous = remove_this.previous;
                   removed = true;
                   break;
               }
               //if i != n, keep iterating through the list
           }
           if(!removed){
               remove_this = remove_this.next;
           }
           i++; 
       }
   }

請注意：如果您在if else情況下首先分配head或tail ， if else此時remove_this為空，因此將head或tail設為null並且打印程序未打印任何內容。