在Java中如何实现双向链表?
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发布时间：2022-04-21 17:35
共1个回答
热心网友
时间：2022-04-11 15:32
双向链表：就是有双向指针，即双向的链域。
链结点的结构：
┌────┬────┬────────┐
│ data │ next │ previous │
└────┴────┴────────┘
双向链表不必是双端链表(持有对最后一个链结点的引用)，双端链表插入时是双向的。
有两条链：一条从头到尾，一条从尾到头，删除遍历时也是双向的。
/**
* 双向链表
*/
public class DoublyLinkedList<t> {
private Link<t> head; //首结点
private Link<t> rear; //尾部指针
public DoublyLinkedList() { }
public T peekHead() {
if (head != null) {
return head.data;
}
return null;
}
public boolean isEmpty() {
return head == null;
}
public void insertFirst(T data) {// 插入 到 链头
Link<t> newLink = new Link<t>(data);
if (isEmpty()) {//为空时，第1次插入的新结点为尾结点
rear = newLink;
} else {
head.previous = newLink; //旧头结点的上结点等于新结点
}
newLink.next = head; //新结点的下结点旧头结点
head = newLink; //赋值后，头结点的下结点是旧头结点，上结点null
}
public void insertLast(T data) {//在链尾 插入
Link<t> newLink = new Link<t>(data);
if (isEmpty()) {
head = newLink;
} else {
rear.next = newLink;
}
newLink.previous = rear;
rear = newLink; //赋值后，尾结点的上结点是旧尾结点，下结点null
}
public T deleteHead() {//删除 链头
if (isEmpty()) return null;
Link<t> temp = head;
head = head.next; //变更首结点，为下一结点
if (head != null) {
head.previous = null;
} else {
rear = null;
}
return temp.data;
}
public T deleteRear() {//删除 链尾
if (isEmpty()) return null;
Link<t> temp = rear;
rear = rear.previous; //变更尾结点，为上一结点
if (rear != null) {
rear.next = null;
} else {
head = null;
}
return temp.data;
}
public T find(T t) {//从头到尾find
if (isEmpty()) {
return null;
}
Link<t> find = head;
while (find != null) {
if (!find.data.equals(t)) {
find = find.next;
} else {
break;
}
}
if (find == null) {
return null;
}
return find.data;
}
public T delete(T t) {
if (isEmpty()) {
return null;
}
Link<t> current = head;
while (!current.data.equals(t)) {
current = current.next;
if (current == null) {
return null;
}
}
if (current == head) {
head = head.next;
if (head != null) {
head.previous = null;
}
} else if (current == rear) {
rear = rear.previous;
if (rear != null) {
rear.next = null;
}
} else {
//中间的非两端的结点，要移除current
current.next.previous = current.previous;
current.previous.next = current.next;
}
return current.data;
}
public boolean insertAfter(T key, T data) {//插入在key之后, key不存在return false
if (isEmpty()) {
return false;
}
Link<t> current = head;
while (!current.data.equals(key)) {
current = current.next;
if (current == null) {
return false;
}
}
Link<t> newLink = new Link<t>(data);
if (current == rear) {
rear = newLink;
} else {
newLink.next = current.next;
current.next.previous = newLink;
}
current.next = newLink;
newLink.previous = current;
return true;
}
public void displayList4Head() {//从头开始遍历
System.out.println("List (first-->last):");
Link<t> current = head;
while (current != null) {
current.displayLink();
current = current.next;
}
}
public void displayList4Rear() {//从尾开始遍历
System.out.println("List (last-->first):");
Link<t> current = rear;
while (current != null) {
current.displayLink();
current = current.previous;
}
}
class Link<t> {//链结点
T data; //数据域
Link<t> next; //后继指针，结点 链域
Link<t> previous; //前驱指针，结点 链域
Link(T data) {
this.data = data;
}
void displayLink() {
System.out.println("the data is " + data.toString());
}
}
public static void main(String[] args) {
DoublyLinkedList<integer> list = new DoublyLinkedList<integer>();
list.insertLast(1);
list.insertFirst(2);
list.insertLast(3);
list.insertFirst(4);
list.insertLast(5);
list.displayList4Head();
Integer deleteHead = list.deleteHead();
System.out.println("deleteHead:" + deleteHead);
list.displayList4Head();
Integer deleteRear = list.deleteRear();
System.out.println("deleteRear:" + deleteRear);
list.displayList4Rear();
System.out.println("find:" + list.find(6));
System.out.println("find:" + list.find(3));
System.out.println("delete find:" + list.delete(6));
System.out.println("delete find:" + list.delete(1));
list.displayList4Head();
System.out.println("----在指定key后插入----");
list.insertAfter(2, 8);
list.insertAfter(2, 9);
list.insertAfter(9, 10);
list.displayList4Head();
}
}
