将项目添加到链表的末尾

Question

I'm studying for an exam, and this is a problem from an old test:

We have a singly linked list with a list head with the following declaration:

class Node {
    Object data;
    Node next;
    Node(Object d,Node n) {
        data = d;
        next = n;
    }
}

Write a method void addLast(Node header, Object x) that adds x at the end of the list.

I know that if I actually had something like:

LinkedList someList = new LinkedList();

I could just add items to the end by doing:

list.addLast(x);

But how can I do it here?

Answer 1

class Node {
    Object data;
    Node next;
    Node(Object d,Node n) {
        data = d ;
        next = n ;
       }

   public static Node addLast(Node header, Object x) {
       // save the reference to the header so we can return it.
       Node ret = header;

       // check base case, header is null.
       if (header == null) {
           return new Node(x, null);
       }

       // loop until we find the end of the list
       while ((header.next != null)) {
           header = header.next;
       }

       // set the new node to the Object x, next will be null.
       header.next = new Node(x, null);
       return ret;
   }
}

Answer 2

You want to navigate through the entire linked list using a loop and checking the "next" value for each node. The last node will be the one whose next value is null. Simply make this node's next value a new node which you create with the input data.

node temp = first; // starts with the first node.

    while (temp.next != null)
    {
       temp = temp.next;
    }

temp.next = new Node(header, x);

That's the basic idea. This is of course, pseudo code, but it should be simple enough to implement.

Answer 3

public static Node insertNodeAtTail(Node head,Object data) {
               Node node = new Node(data);
                 node.next = null;
                if (head == null){
                    return node;
                }
                else{
                    Node temp = head;
                    while(temp.next != null){
                        temp = temp.next;
                    }
                    temp.next = node; 
                    return head;
                }        
    }

Answer 4

If you keep track of the tail node, you don't need to loop through every element in the list.

Just update the tail to point to the new node:

AddValueToListEnd(value) {
    var node = new Node(value);

    if(!this.head) {
        this.head = node;
        this.tail = node;
    } else {
        this.tail.next = node; //point old tail to new node
    }
        
    this.tail = node; //now set the new node as the new tail
}

In plain English:

• Create a new node with the given value
• If the list is empty, point head and tail to the new node
• If the list is not empty, set the old tail.next to be the new node
• In either case, update the tail pointer to be the new node

Answer 5

Here is a partial solution to your linked list class, I have left the rest of the implementation to you, and also left the good suggestion to add a tail node as part of the linked list to you as well.

The node file :

public class Node 
{
    private Object data; 
    private Node next; 

    public Node(Object d) 
    { 
        data = d ;
        next = null;
    }

    public Object GetItem()
    {
        return data;
    }

    public Node GetNext()
    {
        return next;
    }

    public void SetNext(Node toAppend)
    {
        next = toAppend;
    }
}

And here is a Linked List file :

public class LL
{
    private Node head;

    public LL()
    {
        head = null;
    }

    public void AddToEnd(String x)
    {
        Node current = head;

        // as you mentioned, this is the base case
        if(current == null) {
            head = new Node(x);
            head.SetNext(null);
        }

        // you should understand this part thoroughly :
        // this is the code that traverses the list.
        // the germane thing to see is that when the 
        // link to the next node is null, we are at the 
        // end of the list.
        else {
            while(current.GetNext() != null)
                current = current.GetNext();

            // add new node at the end
        Node toAppend = new Node(x);
            current.SetNext(toAppend);
        }
    }
}

Answer 6

The above programs might give you NullPointerException. This is an easier way to add an element to the end of linkedList.

public class LinkedList {
    Node head;

    public static class Node{
        int data;
        Node next;

        Node(int item){
            data = item;
            next = null;
        }
    }
    public static void main(String args[]){
        LinkedList ll = new LinkedList();
        ll.head = new Node(1);
        Node second = new Node(2);
        Node third = new Node(3);
        Node fourth = new Node(4);

        ll.head.next = second;
        second.next = third;
        third.next = fourth;
        fourth.next = null;

        ll.printList();
        System.out.println("Add element 100 to the last");
        ll.addLast(100);
        ll.printList();
    }
    public void printList(){
        Node t = head;
        while(n != null){
            System.out.println(t.data);
            t = t.next;
        }

    }

    public void addLast(int item){
        Node new_item = new Node(item);
        if(head == null){
            head = new_item;
            return;
        }
        new_item.next = null;

        Node last = head;
        Node temp = null;
        while(last != null){
            if(last != null)
                temp = last;
            last = last.next;
        }   

        temp.next = new_item;   
        return;
    }
}

Answer 7

The addLast() needs some optimisation as the while loop inside addLast() has O(n) complexity. Below is my implementation of LinkedList. Run the code with ll.addLastx(i) once and run it with ll.addLast(i) again , you can see their is a lot of difference in processing time of addLastx() with addLast().

Node.java

package in.datastructure.java.LinkedList;

/**
* Created by abhishek.panda on 07/07/17.
*/
public final class Node {
    int data;
    Node next;

    Node (int data){
       this.data = data;
    }
    public String toString(){
      return this.data+"--"+ this.next;
    }
}

LinkedList.java

package in.datastructure.java.LinkedList;
import java.util.ArrayList;
import java.util.Date;

public class LinkedList {

    Node head;
    Node lastx;
    /**
     * @description To append node at end looping all nodes from head
     * @param data
     */
    public void addLast(int data){

        if(head == null){
            head = new Node(data);
            return;
        }
        Node last = head;
        while(last.next != null) {
            last = last.next;
        }
        last.next = new Node(data);
    }


    /**
     * @description This keep track on last node and append to it
     * @param data
     */
    public void addLastx(int data){

        if(head == null){
            head = new Node(data);
            lastx = head;
            return;
        }
        if(lastx.next == null){
            lastx.next = new Node(data);
            lastx = lastx.next;
        }

    }

    public String toString(){
        ArrayList<Integer> arrayList = new ArrayList<Integer>(10);
        Node current = head;
        while(current.next != null) {
            arrayList.add(current.data);
            current = current.next;
        }
        if(current.next == null) {
            arrayList.add(current.data);
        }
        return arrayList.toString();
    }


    public static void main(String[] args) {
        LinkedList ll = new LinkedList();
        /**
         * @description Checking the code optimization of append code
         */
        Date startTime = new Date();
        for (int i = 0 ; i < 100000 ; i++){
            ll.addLastx(i);
        }
        Date endTime = new Date();
        System.out.println("To total processing time : " + (endTime.getTime()-startTime.getTime()));
        System.out.println(ll.toString());
    }

}

Answer 8

循环到链表的最后一个元素，其中 next 指针指向 null 然后修改 next 指针指向一个新节点，该节点具有 data=object 和 next pointer = null

Answer 9

Here's a hint, you have a graph of nodes in the linked list, and you always keep a reference to head which is the first node in the linkedList.
next points to the next node in the linkedlist, so when next is null you are at the end of the list.