java数据结构-链表实现队列
删除与添加只对链表的头部操作复杂度是O(1),如果对链表尾部操作复杂度为O(n)
队列需要在线性结构的一端插入元素在另一端删除删除元素,所以需要在两端同时操作,因此有一端的复杂度为O(n)
链表在头不管删除与插入比较方便,需要用head记录头元素,用tail记录尾节点。
尾巴tail插入元素,头head删除元素相对比较容易
public interface Queue<E> {
int getSize();
boolean isEmpty();
void enqueue(E e);
E dequeue();
E getFront();
}
public class LinkedListQueue<E> implements Queue<E> {
private class Node{
public E e;
public Node next;
public Node(E e, Node next){
this.e = e;
this.next = next;
}
public Node(E e){
this(e, null);
}
public Node(){
this(null, null);
}
@Override
public String toString(){
return e.toString();
}
}
private Node head, tail;
private int size;
public LinkedListQueue(){
head = null;
tail = null;
size = 0;
}
@Override
public int getSize(){
return size;
}
@Override
public boolean isEmpty(){
return size == 0;
}
@Override
public void enqueue(E e){
if(tail == null){
tail = new Node(e);
head = tail;
}
else{
tail.next = new Node(e);
tail = tail.next;
}
size ++;
}
@Override
public E dequeue(){
if(isEmpty())
throw new IllegalArgumentException("Cannot dequeue from an empty queue.");
Node retNode = head;
head = head.next;
retNode.next = null;
if(head == null)
tail = null;
size --;
return retNode.e;
}
@Override
public E getFront(){
if(isEmpty())
throw new IllegalArgumentException("Queue is empty.");
return head.e;
}
@Override
public String toString(){
StringBuilder res = new StringBuilder();
res.append("Queue: front ");
Node cur = head;
while(cur != null) {
res.append(cur + "->");
cur = cur.next;
}
res.append("NULL tail");
return res.toString();
}
public static void main(String[] args){
LinkedListQueue<Integer> queue = new LinkedListQueue<>();
for(int i = 0 ; i < 10 ; i ++){
queue.enqueue(i);
System.out.println(queue);
if(i % 3 == 2){
queue.dequeue();
System.out.println(queue);
}
}
}
}
性能比较
import java.util.Random;
public class Main {
// 测试使用q运行opCount个enqueueu和dequeue操作所需要的时间,单位:秒
private static double testQueue(Queue<Integer> q, int opCount){
long startTime = System.nanoTime();
Random random = new Random();
for(int i = 0 ; i < opCount ; i ++)
q.enqueue(random.nextInt(Integer.MAX_VALUE));
for(int i = 0 ; i < opCount ; i ++)
q.dequeue();
long endTime = System.nanoTime();
return (endTime - startTime) / 1000000000.0;
}
public static void main(String[] args) {
int opCount = 100000;
ArrayQueue<Integer> arrayQueue = new ArrayQueue<>();
double time1 = testQueue(arrayQueue, opCount);
System.out.println("ArrayQueue, time: " + time1 + " s");
LoopQueue<Integer> loopQueue = new LoopQueue<>();
double time2 = testQueue(loopQueue, opCount);
System.out.println("LoopQueue, time: " + time2 + " s");
LinkedListQueue<Integer> linkedListQueue = new LinkedListQueue<>();
double time3 = testQueue(linkedListQueue, opCount);
System.out.println("LinkedListQueue, time: " + time3 + " s");
}
}