数据结构 - 链表(双向链表学习)

在这里插入图片描述

程序里有几个注释的地方还是需要特别注意的,根据单链表自己的逻辑来写一遍双向链表,可能有几个细节处理的不到位。

package DataStructures.LinkedList;

public class DoubleLinkedListDemo {
    public static void main(String []args){

        System.out.println("双向链表的测试");
        //先创建节点
        HeroNode2 hero1 = new HeroNode2(1, "宋江", "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();
        //添加节点
        doubleLinkedList.add(hero1);
        doubleLinkedList.add(hero2);
        doubleLinkedList.add(hero3);
        doubleLinkedList.add(hero4);

        doubleLinkedList.list();

        HeroNode2 newHeroNode = new HeroNode2(4,"公孙胜","入云龙");

        doubleLinkedList.updata(newHeroNode);
        doubleLinkedList.list();
        doubleLinkedList.delete(1);
        doubleLinkedList.list();
    }
}
class DoubleLinkedList{
    private HeroNode2 head = new HeroNode2(0,"","");

    public HeroNode2 getHead() {
        return head;
    }

    //删除一个节点
    //说明
    //对于双向链表,直接找到这个节点,进行自我删除
    public void delete(int no){
        if (head.next == null){
            System.out.println("链表为空,无法删除");
        }
        //这里要加next,因为0是不能删除的
        HeroNode2 temp = head.next;
        boolean flag = false;
        //这里现在是用temp本身来判断
        while (temp != null){
            if (temp.no == no){
                flag = true;
                break;
            }
            temp = temp.next;
        }

        if (flag){
                temp.pre.next = temp.next;
            //注意::这里会有问题???
            //因为如果删除的节点是最后一个节点,下面这句话加上了,会出现空指针异常
            if (temp.next != null) {
                temp.next.pre = temp.pre;
            }
            System.out.println("节点删除成功\n");
        }else {
            System.out.printf("没找到编号为 %d 的节点!删除失败\n",no);
        }
    }
    //修改一个节点的内容
    public void updata(HeroNode2 heroNode){
        if (head.next == null){
            System.out.println("链表为空位,无法修改");
            return;
        }
        //这里要用next,因为更改的时候头结点不能找
        HeroNode2 temp = head.next;
        boolean flag = false;

        while(true){
            if (temp == null){
                break;
            }
            if (temp.no == heroNode.no){
                flag = true;
                break;
            }
            temp = temp.next;
        }

        if (flag){
            temp.name = heroNode.name;
            temp.nickname = heroNode.nickname;
        }else {
            System.out.printf("没找到 编号为 %d 的节点\n",heroNode.no);
        }

    }

    //添加节点

    public void add(HeroNode2 newHeroNode){
        HeroNode2 temp = head;

        while(true){
            if (temp.next == null){
                break;
            }
            temp = temp.next;
        }
        //while退出后temp就是链表最后节点
        //形成双向链表
        temp.next = newHeroNode;
        newHeroNode.pre = temp;
    }
    //遍历输出
    public void list(){

        if (head.next == null){
            System.out.println("链表为空");
            return;
        }
        //这里要加next要不然打印会把头结点打印出来
        HeroNode2 temp = head.next;

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

}


class HeroNode2{
    public int no;
    public String name;
    public String nickname;
    public HeroNode2 next; //指向下一个节点
    public HeroNode2 pre; //指向前一个节点

    public HeroNode2(int no, String name, String nickname) {
        this.no = no;
        this.name = name;
        this.nickname = nickname;
    }

    @Override
    public String toString() {
        return "HeroNode2{" +
                "no=" + no +
                ", name='" + name + '\'' +
                ", nickname='" + nickname + '\'' +
                '}';
    }
}

加一个添加的时候根据no排序

//添加节点并根据no有序
    public void addByOrder(HeroNode2 newHeroNode){
        HeroNode2 temp = head;
        boolean flag = false;

        while (true){
            if (temp.next == null){
                break;
            }

            if (temp.next.no > newHeroNode.no){
                break;
            }else if (temp.next.no == newHeroNode.no){
                flag = true;
                break;
            }
            temp = temp.next;
        }

        if (flag){
            System.out.printf("编号 %d 已存在,不能添加\n",newHeroNode.no);
        }else {
    //这里注意不要写反了,先写if这里,再写剩下两个,要不然就形成了一个环状,这里找了好久bug
            if (temp.next != null) {
                temp.next.pre = newHeroNode;
                newHeroNode.next = temp.next;
            }
            temp.next = newHeroNode;
            newHeroNode.pre = temp;

        }
    }

测试

doubleLinkedList.addByOrder(hero1);
doubleLinkedList.addByOrder(hero4);
doubleLinkedList.addByOrder(hero2);
doubleLinkedList.addByOrder(hero3);

doubleLinkedList.list();
结果:

双向链表的测试
HeroNode2{no=1, name='宋江', nickname='及时雨'}
HeroNode2{no=2, name='卢俊义', nickname='玉麒麟'}
HeroNode2{no=3, name='吴用', nickname='智多星'}
HeroNode2{no=4, name='林冲', nickname='豹子头'}
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转载自blog.csdn.net/weixin_43736084/article/details/101992729