创建n个节点的双向链表并计算节点数的Java程序

1 简介

在此程序中,我们将创建一个双向链表,并计算列表中存在的节点数。要对节点进行计数,我们通过将计数器加1来遍历列表。

双向链表上方显示的节点数为5。

2 算法思路

  • 定义一个Node类,该类代表列表中的一个节点。它具有三个属性:数据,前一个将指向上一个节点,下一个将指向下一个节点。
  • 定义另一个用于创建双向链接列表的类,它具有两个节点:head和tail。最初,头和尾将指向null。
  • addNode()将节点添加到列表中:
    • 它首先检查head是否为空,然后将节点插入为head。
    • 头部和尾部都将指向一个新添加的节点。
    • 头的前一个指针将指向空,而尾的下一个指针将指向空。
    • 如果head不为null,则新节点将插入列表的末尾,以使新节点的前一个指针指向尾。
    • 新的节点将成为新的尾巴。尾巴的下一个指针将指向null。
  • 定义一个变量计数器和新临时节点,该临时节点将指向头节点。
  • 通过使当前节点指向列表中的下一个节点,直到当前点为空,遍历列表以对节点进行计数。
  • 将计数器加1。
  • 定义一个新节点“current”,该节点将指向头部。
  • 打印current.data直到current指向null。
  • 当前将在每次迭代中指向列表中的下一个节点。

3 程序实现

/**
 * 一点教程网: http://www.yiidian.com
 */
public class CountList {  
  
    //Represent a node of the doubly linked list  
  
    class Node{  
        int data;  
        Node previous;  
        Node next;  
  
        public Node(int data) {  
            this.data = data;  
        }  
    }  
  
    //Represent the head and tail of the doubly linked list  
    Node head, tail = null;  
  
    //addNode() will add a node to the list  
    public void addNode(int data) {  
        //Create a new node  
        Node newNode = new Node(data);  
  
        //If list is empty  
        if(head == null) {  
            //Both head and tail will point to newNode  
            head = tail = newNode;  
            //head's previous will point to null  
            head.previous = null;  
            //tail's next will point to null, as it is the last node of the list  
            tail.next = null;  
        }  
        else {  
            //newNode will be added after tail such that tail's next will point to newNode  
            tail.next = newNode;  
            //newNode's previous will point to tail  
            newNode.previous = tail;  
            //newNode will become new tail  
            tail = newNode;  
            //As it is last node, tail's next will point to null  
            tail.next = null;  
        }  
    }  
  
    //countNodes() will count the nodes present in the list  
    public int countNodes() {  
        int counter = 0;  
        //Node current will point to head  
        Node current = head;  
  
        while(current != null) {  
            //Increment the counter by 1 for each node  
            counter++;  
            current = current.next;  
        }  
        return counter;  
    }  
  
    //display() will print out the elements of the list  
    public void display() {  
        //Node current will point to head  
        Node current = head;  
        if(head == null) {  
            System.out.println("List is empty");  
            return;  
        }  
        System.out.println("Nodes of doubly linked list: ");  
        while(current != null) {  
            //Prints each node by incrementing the pointer.  
  
            System.out.print(current.data + " ");  
            current = current.next;  
        }  
    }  
  
    public static void main(String[] args) {  
  
        CountList dList = new CountList();  
        //Add nodes to the list  
        dList.addNode(1);  
        dList.addNode(2);  
        dList.addNode(3);  
        dList.addNode(4);  
        dList.addNode(5);  
  
        //Displays the nodes present in the list  
        dList.display();  
  
        //Counts the nodes present in the given list  
        System.out.println("\nCount of nodes present in the list: " + dList.countNodes());  
    }  
}  

输出结果为:

Nodes of doubly linked list: 
1 2 3 4 5 
Count of nodes present in the list: 5

 

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