从链表java中删除元素

Question

HI :)我有一个关于链接列表的程序，我们应该能够删除两个数字，如果它们是相同的...我知道如何从一开始就这样做但是如果它们是如何删除两个数字链表的中间?? 所有3个一起运行Heres我的数字计划

import java.util.Scanner;

public class Numbers {

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner reader = new Scanner (System.in);
        LinkedList link=new LinkedList();
        LinkedList link2= new LinkedList();
        System.out.println("Enter in 5 numbers to put in your list");
        int num1, num2, num3, num4, num5;
        num1 = reader.nextInt();
        link.addToStart(num1);
        num2 = reader.nextInt();
        link.addToStart(num2);
        num3 = reader.nextInt();
        link.addToStart(num3);
        num4 = reader.nextInt();
        link.addToStart(num4);
        num5 = reader.nextInt();
        link.addToStart(num5);

        link2.addToStart(num5);
        link2.addToStart(num4);
        link2.addToStart(num3);
        link2.addToStart(num2);
        link2.addToStart(num1);

        System.out.println("The size of the linked list is " + link.size());

        System.out.print("Here is the list ");
        link2.outputList();
        System.out.println();
        System.out.print("Here is the list in reverse order ");
        link.outputList( );
        System.out.println();

        if (num1==num2){
             link2.deleteHeadNode(num1);
             link2.deleteHeadNode(num2);
             System.out.println("Here is the list with the removed numbers");
            link2.outputList();
            System.out.println();
            System.out.println("Here is its size");
            System.out.println(link2.size());
        }
        else if (num2==num3){
            link2.deleteHeadNode(num2);
            link2.deleteHeadNode(num3);
            System.out.println("Here is the list with the removed numbers");
           link2.outputList();
           System.out.println();
           System.out.println("Here is its size");
           System.out.println(link2.size());
       }
    }

}

这是节点程序

public class Node1
{
    private Object item;
    private int count;
    private Node1 link;

    public Node1( )
    {
        link = null;
       item = null;
        count = 0;
    }

    public Node1(int num, int newCount, Node1 linkValue)
    {
        setData(num, newCount);
        link = linkValue;
    }

    public void setData(int num, int newCount)
    {
        item = num;
        count = newCount;
    }

    public void setLink(Node1 newLink)
    {
        link = newLink;
    }

    public Object getItem( )
    {
        return  item;
    }

    public int getCount( )
    {
        return count;
    }

    public Node1 getLink( )
    {
        return link;
    }
}

这里是链接列表程序

public class LinkedList
{
    private Node1 head;

    public LinkedList( )
    {
        head = null;
    }

    /**
     Adds a node at the start of the list with the specified data.
     The added node will be the first node in the list.
    */
    public void addToStart(int num)
    {
        head = new Node1(num, num, head);
    }

    /**
     Removes the head node and returns true if the list contains at least
     one node. Returns false if the list is empty.
     * @param num1 
    */
    public boolean deleteHeadNode(int num1 )
    {
        if (head != null)
        {
            head = head.getLink( );
            return true;
        }
        else
            return false;
    }

    /**
     Returns the number of nodes in the list.
    */
    public int size( )
    {
        int count = 0;
        Node1 position = head;

        while (position != null)
        {
            count++;
            position = position.getLink( );
        }
        return count;
    }

    public boolean contains(String item)
    {
        return (find(item) != null);
    }

    /**
     Finds the first node containing the target item, and returns a
     reference to that node. If target is not in the list, null is returned.
    */
    private Node1 find(String target)
    {
        Node1 position = head;
        Object itemAtPosition;
        while (position != null)
        {
            itemAtPosition = position.getItem( );
            if (itemAtPosition.equals(target))
                return position;
            position = position.getLink( );
        }
        return null; //target was not found
    }

    public void outputList( )
    {
        Node1 position = head;
        while (position != null)
        {
            System.out.print(position.getItem( ) + " ");
            position = position.getLink( );
        }
    }

    public boolean isEmpty( )
    {
        return (head == null);
    }

    public void clear( )
    {
        head = null;
    }

}

Answer 1

要删除链接列表中间的项目，请将上一项的“链接”指针设置为要删除的对象的“链接”指针。 例如，您可以将以下内容添加到LinkedList类：

public void removeNode(Node previousNode, Node nodeToRemove) {
  if (previousNode != null) {
    previousNode.setLink(nodeToRemove.getLink());
  }
}

要更好地考虑这个，请画一幅画。

 N1 -> N2 -> N3 -> N4

N1的“链接”是N2等。如果你想删除N2，只需将N1的“链接”设置为N3。

 N1 -> N3 -> N4

Answer 2

执行蛮力查找的一种方法。

• 对于您搜索的每个元素，是否在列表中重复。
• 如果是，则将其删除
• 然后去下一个。

正如您可能看到这三个步骤可能很容易编码，这里的重点是首先要了解它们是否符合您的要求。

这是这三点的伪代码：

forEach( Element a : inList ) do
    // e is the element we want to find repeated.
    forEach( Element b : inList ) do  
         // b is the element in the list.
         if( a == b ) then // repeated
             inList.remove( a ) 
             break;
         endIf
     endFor
 endFor

此方法将允许您删除所有重复的元素。

只记得删除一个项目，你必须确保你没有丢失它的参考。 所以如果你有：

n1 -> n2 -> n3

在某些时候你必须有n1和n2指向n3 （这样n1保持参考n2有）

n1 -> n3  n2 ->n3

然后删除n2，它离开你：

n1 -> n3

现在，如何使用您的特定数据结构进行编码是您必须执行的任务;）

Answer 3

这是一种方法。

public void delete(int item)
{
        while(head.data==item)              //For deleting head
    {

        head=head.link;

    }
    // For middle elements..................
    Node ptr, save;
    save=head;
    ptr=head.link;
    while(ptr!=null)
    {
        if(ptr.data==item)
        {
        Node next=ptr.link;
        save.link=next;
        ptr=next;
        }
        else
        {
            save=ptr;
            ptr=ptr.link;
        }

}
}