双向链表

 

 

 

 

修改的时候思路和原来的单向链表思路一样

 

 

public class DoubleLinkedListDemo {
//初始化头节点,里面的pre还有next不动
HeroNode2 head = new HeroNode2(0,"","");

public HeroNode2 getHeroNode2(){
return head;
}

public void add(HeroNode2 head){
HeroNode2 temp = head;
while (true) {
if (temp.next == null) {
break;
}temp=temp.next;
}
temp.next = head; //temp.next = newHeadNode
head.pre=temp; //newHeadNode.pre = temp;
}

public void list(){
if (head.next == null) {
System.out.println("链表为空");
return;
}
HeroNode2 temp = head.next;
while (true) {
if (temp == null) {
break;
}
System.out.println(temp);
temp=temp.next;
}
}
public void update(HeroNode2 heroNode2){
if (heroNode2.next == null) {
System.out.println("链表为空");
return;
}
HeroNode2 temp = heroNode2.next;
boolean flag =false;
while (true) {
if (temp.next == null) {
flag=true;
break;
}
if (temp.no == heroNode2.no) {
flag=true;
break;
}
temp=temp.next;
}
if (flag) {
temp.name = heroNode2.name;
temp.nickName =heroNode2.nickName;
}else {
System.out.println("没有找到这个节点的编号"+heroNode2.no);
}
}

/**
* 双向链表中删除一个节点
* 双向链表可以直接找到删除的这个节点
* @param no
*/
public void del(int no){
if (head.next == null) {
System.out.println("空链表");
return;
}
HeroNode2 temp = head.next;//单链表是待删除的前一个,双链表直接从第一个节点开始
boolean flag=false;
while (true) {
if (temp == null) {
break;
}
if (temp.no == no) {
flag=true;
break;
}
temp=temp.next;
}
if (flag) {
temp.pre.next=temp.next;
//代码有问题 如果是最后一个节点就不需要执行下面这句话 否则空指针
if (temp.next != null) {
temp.next.pre=temp.pre;
}
}else {
System.out.println("要删除的不存在");
}
}

}
class HeroNode2{
//属性和节点信息,每个heronode就算一个节点
public int no;
public String name;
public String nickName;
public int val;
public HeroNode2 next;
public HeroNode2 pre;

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

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

-----------------------------------------------------------------------------
修改后
public class DoubleLinkedListDemo {

public static void main(String[] args) {
DoubleLinkedList doubleLinkedList = new DoubleLinkedList();
HeroNode2 heroNode1 = new HeroNode2(1, "松江", "及时雨");
HeroNode2 heroNode2 = new HeroNode2(2, "卢俊义", "玉麒麟");
HeroNode2 heroNode3 = new HeroNode2(3, "吴用", "智多星");
HeroNode2 heroNode4 = new HeroNode2(4, "林冲", "豹子头");
doubleLinkedList.add(heroNode1);
doubleLinkedList.add(heroNode2);
doubleLinkedList.add(heroNode3);
doubleLinkedList.add(heroNode4);

doubleLinkedList.list();

System.out.println("---------------------------------");
HeroNode2 newheroNode4 = new HeroNode2(4, "公孙胜", "入云龙");
doubleLinkedList.update(newheroNode4);
doubleLinkedList.list();

doubleLinkedList.del(3);
doubleLinkedList.list();



}
}

class DoubleLinkedList {
//初始化头节点,里面的pre还有next不动
private HeroNode2 head = new HeroNode2(0, "", "");

public HeroNode2 getHead() {
return head;
}

public void add(HeroNode2 heroNode2) {
HeroNode2 temp = head;
while (true) {
if (temp.next == null) {
break;
}
temp = temp.next;
}
temp.next = heroNode2; //temp.next = newHeadNode
heroNode2.pre = temp; //newHeadNode.pre = temp;
}

public void list() {
if (head.next == null) {
System.out.println("链表为空");
return;
}
HeroNode2 temp = head.next;
while (true) {
if (temp == null) {
break;
}
System.out.println(temp);
temp = temp.next;
}
}

public void update(HeroNode2 heroNode2) {
if (head.next == null) {
System.out.println("链表为空");
return;
}
HeroNode2 temp = head.next;
boolean flag = false;
while (true) {
if (temp == null) {
break;
}
if (temp.no == heroNode2.no) {
flag = true;
break;
}
temp = temp.next;
}
if (flag) {
temp.name = heroNode2.name;
temp.nickName = heroNode2.nickName;
} else {
System.out.println("没有找到这个节点的编号" + heroNode2.no);
}
}

/**
* 双向链表中删除一个节点
* 双向链表可以直接找到删除的这个节点
*
* @param no
*/
public void del(int no) {
if (head.next == null) {
System.out.println("空链表");
return;
}
HeroNode2 temp = head.next;//单链表是待删除的前一个,双链表直接从第一个节点开始
boolean flag = false;
while (true) {
if (temp == null) {
break;
}
if (temp.no == no) {
flag = true;
break;
}
temp = temp.next;
}
if (flag) {
temp.pre.next = temp.next;
//代码有问题 如果是最后一个节点就不需要执行下面这句话 否则空指针
if (temp.next != null) {
temp.next.pre = temp.pre;
}
} else {
System.out.println("要删除的不存在");
}
}
}

class HeroNode2{
//属性和节点信息,每个heronode就算一个节点
public int no;
public String name;
public String nickName;
public HeroNode2 next;
public HeroNode2 pre;

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

// public HeroNode2(final int no, final String name, final String nickName, final HeroNode2 next, final HeroNode2 pre) {
// this.no = no;
// this.name = name;
// this.nickName = nickName;
// this.next = next;
// this.pre = pre;
// }


@Override
public String toString() {
return "HeroNode2{" +
"no=" + no +
", name='" + name + '\'' +
", nickName='" + nickName + '\'' +
'}';
}
}
 
posted @ 2021-11-05 10:14  lamda表达式先驱  阅读(36)  评论(0)    收藏  举报