欢迎访问悦橙教程(wld5.com),关注java教程。悦橙教程  java问答|  每日更新
页面导航 : > > 文章正文

java双向链表实现代码,java实现代码,一个LinkedList

来源: javaer 分享于  点击 42108 次 点评:196

java双向链表实现代码,java实现代码,一个LinkedList


一个LinkedList的实现,非java util 的实现,没有用java util容器,纯实现过程,主要看算法。

[Java]代码

/* * To change this template, choose Tools | Templates * and open the template in the editor. */package linkedlisttest;/** * * @author Lindily */public class LinkedList<E> {    int size = 0;    Node<E> head = new Node(null, null, null);    public LinkedList() {        head.next = head.previous = head;    }    public void add(E node) {        //核心 循环双向链表        Node<E> newNode = new Node(head.previous, node, head);   //新节点的prev指向头结点的prev 新节点的next指向头结点        newNode.previous.next = newNode;    //调整,新节点的前一个的后一个        newNode.next.previous = newNode;    //调整,新节点的后一个的前一个        size++;    }    public Node get(int index) {        if (index < 0 || index >= size) {            throw new IndexOutOfBoundsException("Index:" + index + ",size:" + size);        }        Node node = head;        if (index < (size >> 1)) {          //index转为二进制带符号向右移动一个bit,相当于index/2            for (int i = 0; i <= index; i++) {  //head是哑元,i<=index当index=0时,返回head.next                node = node.next;           //对头结点进行迭代            }        }else{            for(int i=size;i>index;i--){                node=node.previous;            }        }        return node;    }    public int size() {        return size;    }    public static void main(String[] args) {        LinkedList list = new LinkedList();        list.add(1);        list.add(2);        list.add(3);        for (int i = 0; i < list.size; i++) {            System.out.println(list.get(i).node);        }    }}class Node<E> {    E node;    Node<E> next;    Node<E> previous;    public Node(Node<E> previous, E node, Node<E> next) {        this.node = node;        this.next = next;        this.previous = previous;    }}

[Java]代码

/* * To change this template, choose Tools | Templates * and open the template in the editor. */package linkedlisttest;/** * * @author Lindily */public class SingleLinkedList<T> {    int size = 0;    Node<T> head, tail;    public SingleLinkedList() {        head = tail = null;    }    public void add(T node) {        if (size == 0) {            head = tail = new Node<T>(node, null);        } else {            tail.next = new Node<T>(node, null);            tail = tail.next;        }        size++;    }    public Node<T> get(int index) {        if (index < 0 || index >= size) {            throw new IndexOutOfBoundsException("Index:" + index + ",size:" + size);        }        Node<T> node = head;        for (int i = 0; i < index; i++) {  //当index=0时,i不小于index(零不小于零),于是直接return头结点,与linkedList不同,head不是哑元            node = node.next;           //对头结点进行迭代        }        return node;    }    public int size(){        return size;    }    public static void main(String[] args){        SingleLinkedList list=new SingleLinkedList();        list.add(1);        list.add(2);        list.add(3);        for(int i=0;i<list.size();i++){            System.out.println(list.get(i).node);        }    }}class Node<T> {    T node;    Node<T> next;    public Node(T node, Node next) {        this.node = node;        this.next = next;    }}
相关栏目:

用户点评