【难点】：remove、removeAllKey
一、IList
java
package bagfive;
/**
* Created with IntelliJ IDEA.
* Description:
* User: tangyuxiu
* Date: 2024-08-21
* Time: 20:30
*/
public interface IList<E> {
//头插法
void addFirst(E data);
//尾插法
void addLast(E data);
//任意位置插入,第一个数据节点为0号下标
void addIndex(int pos,E data);
//查找是否包含关键字key是否在单链表当中
boolean contains(E key);
//删除第一次出现关键字为key的节点
void remove(E key);
//删除所有值为key的节点
void removeAllKey(E key);
//得到单链表的长度
int size();
void display();
void clear();
}二、MyLinkedList
java
package bagfive;
/**
* Created with IntelliJ IDEA.
* Description:
* User: tangyuxiu
* Date: 2024-08-21
* Time: 20:30
*/
public class MyLinkedList<E> implements IList<E> {
/* 使用内部类定义双向链表节点 */
private static class ListNode<E> {
E val;
ListNode<E> prev;
ListNode<E> next;
public ListNode(E val) {
this.val = val;
}
}
private ListNode<E> head;
private ListNode<E> last;
@Override
public void addFirst(E data) {
ListNode<E> newNode = new ListNode<>(data);
//1. 如果链表为null
if (this.head == null) {
this.head = this.last = newNode;
return;
}
//2. 链表不为null
newNode.next = this.head;
this.head.prev = newNode;
this.head = newNode;
}
@Override
public void addLast(E data) {
ListNode<E> newNode = new ListNode<>(data);
//1. 如果链表为null
if (this.head == null) {
this.head = this.last = newNode;
return;
}
//2. 链表不为null
this.last.next = newNode;
newNode.prev = this.last;
this.last = newNode;
}
/* 判断add的位置是否合法 */
private boolean addIndexIsLegal(int pos) {
if (pos < 0 || pos > this.size()) {
return false;
}
return true;
}
@Override
public void addIndex(int pos, E data) {
//1. 判断pos位置是否合法
if (!this.addIndexIsLegal(pos)) {
return;
}
//2. pos == 0
if (pos == 0) {
this.addFirst(data);
return;
}
//3. pos == size()
if (pos == this.size()) {
this.addLast(data);
return;
}
//4. 其他位置,不需要找前驱节点了
ListNode<E> newNode = new ListNode<>(data);
ListNode<E> cur = this.head;
for (int i = 0; i < pos; i++) {
cur = cur.next;
}
newNode.next = cur;
cur.prev.next = newNode;
newNode.prev = cur.prev;
cur.prev = newNode;
}
@Override
public boolean contains(E key) {
ListNode<E> cur = this.head;
while (cur != null) {
if (cur.val.equals(key)) {
return true;
}
cur = cur.next;
}
return false;
}
@Override
public void remove(E key) {
// 头，任意，尾 (链表为null进不去循环)
ListNode<E> del = this.head;
while (del != null) {
if (del.val.equals(key)) {
if (this.head == del) {//如果为头节点
if (this.head.next == null) {//如果为头节点，且链表只有一个节点
this.head = this.last = null;
} else {
this.head = this.head.next;
this.head.prev = null;
}
} else {
if (this.last == del) {//如果为尾节点
this.last = this.last.prev;
this.last.next = null;
} else {//其他位置
del.prev.next = del.next;
del.next.prev = del.prev;
}
}
return;
}
del = del.next;
}
}
@Override
public void removeAllKey(E key) {
// 头，只一个节点，任意，尾 (链表为null进不去循环)
ListNode<E> del = this.head;
while (del != null) {
if (del.val.equals(key)) {
if (this.head == del) {//如果为头节点
if (this.head.next == null) {//如果为头节点，且链表只有一个节点
this.head = this.last = null;
} else {
this.head = this.head.next;
this.head.prev = null;
}
} else {
if (this.last == del) {//如果为尾节点
this.last = this.last.prev;
this.last.next = null;
} else {
del.prev.next = del.next;//其他位置
del.next.prev = del.prev;
}
}
}
del = del.next;
}
}
@Override
public int size() {
int count = 0;
ListNode<E> cur = this.head;
while (cur != null) {
count++;
cur = cur.next;
}
return count;
}
@Override
public void display() {
ListNode<E> cur = this.head;
while (cur != null) {
System.out.print(cur.val + " ");
cur = cur.next;
}
System.out.println();
}
@Override
public void clear() {
ListNode<E> cur = this.head;
while (cur != null) {
cur.val = null;
cur = cur.next;
}
this.head = this.last = null;//🙀
}
}