Doubly linked list
Analysis of shortcomings of one-way linked list
- One-way linked list, the search direction can only be one direction, and the two-way linked list can search forward or backward
- The one-way linked list cannot be deleted by itself, it needs to rely on auxiliary nodes, and the two-way linked list can be deleted by itself, so when deleting the one-way linked list node in the previous, you need to find the next node of temp to delete
- If you don't understand singly linked list, please refer to-> Single linked list click here
The schematic diagram of the doubly linked list is as follows:
Analyze the operation idea of traversing, adding, modifying and deleting doubly linked lists
The traversal method is the same as the one-way linked list, except that it can be added forward or backward
(it is added to the end of the doubly linked list by default)
- First find the last node of the doubly linked list
- temp.next points to the new node, temp.next = newHeroNode
- newHerNOde.pre=temp;
The modification idea is the same as the one-way linked list
deletion
- Because it is a two-wire linked list, we can delete a node by ourselves
- Find the node to be deleted, such as temp
- temp.per.next = temp.next, (point next to the node before temp and the node after temp)
- temp.next.pre=temp.pre
Code
package DataType;
public class DoubleLinkedLIstDemo {
public static void main(String[] args){
//测试
System.out.println("双向链表的测试~~~");
//先创建节点
HeroNode2 hero1 = new HeroNode2(1, "松江", "及时雨");
HeroNode2 hero2 = new HeroNode2(2, "卢俊义", "玉麒麟");
HeroNode2 hero3 = new HeroNode2(3, "吴用", "智多星");
HeroNode2 hero4 = new HeroNode2(4, "林冲", "豹子头");
//创建一个双向链表
DoubleLinkedList doubleLinkedList = new DoubleLinkedList();
doubleLinkedList.add(hero1);
doubleLinkedList.add(hero2);
doubleLinkedList.add(hero3);
doubleLinkedList.add(hero4);
doubleLinkedList.list();
//修改测试
HeroNode2 newHeroNode = new HeroNode2(4, "董小宇", "董小宇");
doubleLinkedList.update(newHeroNode);
System.out.println("修改后的链表是~~~");
doubleLinkedList.list();
//删除测试
doubleLinkedList.del(3);
System.out.println("删除后的链表情况~~~");
doubleLinkedList.list();
//测试按照顺序添加
HeroNode2 addHeroNode = new HeroNode2(3, "董大宇", "董大宇");
doubleLinkedList.addByOrder(addHeroNode);
System.out.println("按照顺序后的链表情况~~~");
doubleLinkedList.list();
}
}
//创建一个双向链表的类
class DoubleLinkedList{
///先初始化一个头节点，头节点不要动，固定的位置，不存放具体的数据
private HeroNode2 head = new HeroNode2(0,"","");
//返回头节点
public HeroNode2 getHead(){
return head;
}
//遍历双向链表的方法
public void list(){
//先判断链表是否为空
if(head.next == null){
System.out.println("链表为空！！");
return;
}
//因为头节点，不能动，因此还需要辅助变量来遍历
HeroNode2 temp = head.next;
while (true){
//判断是否到了链表最后
if (temp == null){
break;
}
//输出节点的信息
System.out.println(temp);
//将temp后移
temp = temp.next;
}
}
//添加一个节点，到双向链表的最后
public void add(HeroNode2 heroNode){
//因为head不能动，因此我们需要一个辅助变量 temp
HeroNode2 temp = head;
//遍历链表，找到最后
while (true){
//找到链表的最后
if (temp.next == null){
break;
}
//如果没有找到最后，就将temp后移
temp = temp.next;
}
//当推出while循环时，temp就指向了链表的最后
//构成一个双向链表
temp.next = heroNode;
heroNode.pre = temp;
}
//添加数据，按照顺序来
public void addByOrder(HeroNode2 heroNode){
//还是因为头节点不能动，我们仍然通过辅助指针来帮助
//因为双向链表，因此我们找的temp
HeroNode2 temp = head;
boolean flag = false;//标志添加的编号是否存在，默认为false
while (true){
if (temp.next == null){//说明temp在链表最后
break;//
}
if (temp.next.no > heroNode.no){//位置找到了，就在temp的后面输入
break;
}else if (temp.next.no == heroNode.no){//说明添加数据的编号已经存在了
flag = true;//说明编号存在、
break;
}
temp = temp.next;//后移，相当于遍历
}
//判断flag的值
if (flag){//如果为真，就不能添加，说明编号存在
System.out.printf("准备插入英雄编号%d已经存在，不能加入\n",heroNode.no);
}else{
//插入到链表当中，temp的后面
heroNode.next = temp.next;
temp.next = heroNode;
heroNode.pre = temp.next.pre;
temp.next.pre = heroNode;
}
}
//修改一个节点的内容，可以看到双向链表节点修改和单向链表一样，只是类型不一样，
public void update(HeroNode2 heroNode){
//判断是否为空
if (head.next == null){
System.out.println("链表为空！");
return;
}
//找到需要修改的节点，根据no编号查找
//定义一个temp辅助变量
HeroNode2 temp = head.next;
boolean flag = false;//表示是否找到该节点
while (true){
if (temp == null){
break;//到链表最后了，没有数据了，已经遍历结束了
}
if (temp.no == heroNode.no){
//找到了节点
flag = true;
break;
}
temp = temp.next;
}
//根据flag判断是否找到要修改的节点
if (flag){
temp.name = heroNode.name;
temp.nickName = heroNode.nickName;
}else {//没有找到节点
System.out.printf("没有找到编号%d的节点，不能修改\n",heroNode.no);
}
}
//从双向链表中删除一个节点
//说明
//1.对于双向链表，可以直接找到要删除的节点
//2.找到后自我删除即可
public void del(int no){
//判断当前当前列表是否为空
if (head.next == null){//空链表
System.out.println("链表空，无法删除！");
return;
}
HeroNode2 temp = head.next;//辅助指针
boolean flag = false;//标记是否找到需要删除的节点
while (true){
if (temp == null){//说明遍历结束，已经到最后了
break;
}
if (temp.no == no){
//说明找到了需要删除的节点temp
flag = true;
break;
}
temp = temp.next;//temp后移进行遍历
}
//判断flag
if (flag){
//找到，可以删除
//相当于把需要删除的节点排除出去了
temp.pre.next = temp.next;
if (temp.next != null) {
//代码有弊端，然后是最后一个节点，就不需要执行下面这句话，不然会空指针异
temp.next.pre = temp.pre;
}
}else{
System.out.printf("要删除的%d节点不存在",no);
}
}
}
//定义一个HeroNode2，每个HeroNode对象就是一个节点
class HeroNode2{
public int no;
public String name;
public String nickName;
public HeroNode2 next;//指向下一个节点，默认为null
public HeroNode2 pre;//指向前一个节点，默认为null
//构造器
public HeroNode2(int no,String name,String nickName){
this.no = no;
this.name = name;
this.nickName = nickName;
}
//为了显示方法，我们重新toString
@Override
public String toString() {
return "[HeroNode2{" +
"no=" + no +
", name='" + name + '\'' +
", nickName='" + nickName + "]";
}
}
I am also a beginner in programming, and it is inevitable that there is a mistake in understanding. I hope that after reading, you can comment out the errors, thank you
