多线程模拟题

本文主要是介绍多线程模拟题，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

1. 编写一个程序，启动三个线程，三个线程的 ID 分别是 A，B，C；，每个线程将自己的 ID 值在屏幕上打印 5 遍，打印顺序是 ABCABC…

	package test;
	//编写一个程序，启动三个线程，三个线程的ID分别是A，B，C；，每个线程将自己的ID值在屏幕上打印5遍，打印顺序是ABCABC…
	public class ABCABCABC {
	public static int count=0;
	public static char v ='A';
	public synchronized void A(){
	for(int i=0;i<5;i++){
	while (v!='A'){
	try {
	this.wait();
	} catch (InterruptedException e) {
	e.printStackTrace();
	}
	}
	System.out.println("A");
	v='B';
	notifyAll();
	}
	}
	public synchronized void B(){
	for(int i=0;i<5;i++){
	while (v!='B'){
	try {
	this.wait();
	} catch (InterruptedException e) {
	e.printStackTrace();
	}
	}
	System.out.println("B");
	v='C';
	notifyAll();
	}
	}
	public synchronized void C(){
	for(int i=0;i<5;i++){
	while (v!='C'){
	try {
	this.wait();
	} catch (InterruptedException e) {
	e.printStackTrace();
	}
	}
	System.out.println("C");
	v='A';
	notifyAll();
	}
	}
	public static void main(String[] args) {
	ABCABCABC ab = new ABCABCABC();
	new Thread(()->ab.A()).start();
	new Thread(()->ab.B()).start();
	new Thread(()->ab.C()).start();
	}
	}

   

2. 打印 ABC

	package test;
	// 打印ABC
	public class Main1 {
	static char v ='A';
	public static void main(String[] args) {
	final Main1 main1 = new Main1();
	Thread thread1 = new Thread(() -> main1.A());
	Thread thread2 = new Thread(() -> main1.B());
	Thread thread3= new Thread(() -> main1.C());
	thread1.start();
	thread2.start();
	thread3.start();
	}
	private synchronized void A() {
	while (true){
	if(v == 'A'){
	System.out.println('A');
	v = 'B';
	this.notifyAll();
	}else{
	try {
	this.wait();
	} catch (InterruptedException e) {
	e.printStackTrace();
	}
	}
	}
	}
	private synchronized void B() {
	while (true){
	if(v == 'B'){
	System.out.println('B');
	v = 'C';
	this.notifyAll();
	}else{
	try {
	this.wait();
	} catch (InterruptedException e) {
	e.printStackTrace();
	}
	}
	}
	}
	private synchronized void C() {
	while (true){
	if(v == 'C'){
	System.out.println('C');
	v = 'A';
	this.notifyAll();
	}else{
	try {
	this.wait();
	} catch (InterruptedException e) {
	e.printStackTrace();
	}
	}
	}
	}
	}

   

3. 两个线程轮流打印数字，一直到 100

	package test;
	/**
	* 两个线程轮流打印数字，一直到100
	*
	* Java的wait()、notify()学习：
	* https://blog.csdn.net/boling_cavalry/article/details/77995069
	*/
	public class TakeTurnsPrint {
	public static class TakeTurns {
	private static boolean flag = true;
	private static int count = 0;
	public synchronized void print1() {
	for (int i = 1; i <= 50; i++) {
	while (!flag) {
	try {
	System.out.println("print1: wait before");
	wait();
	System.out.println("print1: wait after");
	} catch (InterruptedException e) {
	}
	}
	System.out.println("print1: " + ++count);
	flag = !flag;
	notifyAll();
	}
	}
	public synchronized void print2() {
	for (int i = 1; i <= 50; i++) {
	while (flag) {
	try {
	System.out.println("print2: wait before");
	wait();
	System.out.println("print2: wait after");
	} catch (InterruptedException e) {
	}
	}
	System.out.println("print2: " + ++count);
	flag = !flag;
	notifyAll();
	}
	}
	}
	public static void main(String[] args){
	TakeTurns takeTurns = new TakeTurns();
	new Thread(()->takeTurns.print1()).start();
	new Thread(()->takeTurns.print2()).start();
	}
	}

   

4. 写两个线程，一个线程打印 1~ 52，另一个线程打印 A~Z，打印顺序是 12A34B…5152Z

	package test;
	public class TakeTurnsPrint2 {
	private boolean flag = true; // 初始化flag状态，确保第一个线程开始执行
	private int count = 1; // 从1开始计数
	public synchronized void printNum() {
	for (int i = 0; i < 26; i++) {
	while (!flag) { // 等待切换到数字打印
	try {
	wait();
	} catch (InterruptedException e) {
	e.printStackTrace();
	}
	}
	System.out.print(count++); System.out.print(count++);
	flag = !flag; // 切换标志
	notifyAll(); // 唤醒其他等待的线程
	}
	}
	public synchronized void printLetter() {
	for (char ch = 'A'; ch <= 'Z'; ch++) { // 直接使用char遍历字母表
	while (flag) { // 等待切换到字母打印
	try {
	wait();
	} catch (InterruptedException e) {
	e.printStackTrace();
	}
	}
	System.out.println(ch);
	flag = !flag; // 切换标志
	notifyAll(); // 唤醒其他等待的线程
	}
	}
	public static void main(String[] args) {
	TakeTurnsPrint2 takeTurnsPrint2 = new TakeTurnsPrint2();
	new Thread(takeTurnsPrint2::printNum).start();
	new Thread(takeTurnsPrint2::printLetter).start();
	}
	}

   

5. 编写 10 个线程，第一个线程从 1 加到 10，第二个线程从 11 加 20… 第十个线程从 91 加到 100，最后再把 10 个线程结果相加。

	package test;
	//编写10个线程，第一个线程从1加到10，第二个线程从11加20…第十个线程从91加到100，最后再把10个线程结果相加。
	public class TenThreadSum {
	public static class SumThread extends Thread{
	int forct = 0; int sum = 0;
	SumThread(int forct){
	this.forct = forct;
	}
	@Override
	public void run() {
	for(int i = 1; i <= 10; i++){
	sum += i + forct * 10;
	}
	System.out.println(getName() + " " + sum);
	}
	}
	public static void main(String[] args) {
	int result = 0;
	for(int i = 0; i < 10; i++){
	SumThread sumThread = new SumThread(i);
	sumThread.start();
	try {
	sumThread.join();
	} catch (InterruptedException e) {
	e.printStackTrace();
	}
	//join()方法被用来确保每个SumThread线程执行完毕并计算出自己的求和结果之后，主线程再收集这些结果进行最终的总和计算
	result = result + sumThread.sum;
	}
	System.out.println("result " + result);
	}
	}

   

6. 创建四个线程，2 个增加，2 个删除，保证数据合法

	package test;
	public class Test1 {
	private int j ;//变量j
	public static void main(String[] args) {
	Test1 t=new Test1();
	Add add=t.new Add();//内部类的实例
	Sub sub=t.new Sub();
	for(int i=0;i<2;i++){//循环两边创建四个线程，两加两减
	Thread t1=new Thread(add);
	t1.start();
	Thread t2=new Thread(sub);
	t2.start();
	}
	}
	private synchronized void add(){
	j++;
	System.out.println(Thread.currentThread().getName()+":"+j);
	}
	private synchronized void sub(){
	j--;
	System.out.println(Thread.currentThread().getName()+":"+j);
	}
	class Add extends Thread{
	@Override
	public void run() {
	for(int i=0;i<10;i++){
	add();
	}
	}
	}
	class Sub extends Thread{
	@Override
	public void run() {
	for (int i=0;i<10;i++){
	sub();
	}
	}
	}
	}

   

7. 三个窗口同时买票

	package test;
	/**
	* 三个窗口同时卖票
	*/
	/**
	* 票
	*/
	class Ticket {
	private int count = 1;
	public void sale() {
	while (true) {
	synchronized (this) {
	if (count > 200) {
	System.out.println("票已经卖完啦");
	break;
	} else {
	System.out.println(Thread.currentThread().getName() + "卖的第 " + count++ + " 张票");
	}
	try {
	Thread.sleep(200);
	} catch (InterruptedException e) {
	e.printStackTrace();
	}
	}
	}
	}
	}
	/**
	* 售票窗口
	*/
	class SaleWindows extends Thread {
	private Ticket ticket;
	public SaleWindows(String name, Ticket ticket) {
	super(name);
	this.ticket = ticket;
	}
	@Override
	public void run() {
	super.run();
	ticket.sale();
	}
	}
	public class TicketDemo {
	public static void main(String[] args) {
	Ticket ticket = new Ticket();
	SaleWindows windows1 = new SaleWindows("窗口1", ticket);
	SaleWindows windows2 = new SaleWindows("窗口2", ticket);
	SaleWindows windows3 = new SaleWindows("窗口3", ticket);
	windows1.start();
	windows2.start();
	windows3.start();
	}
	}

   

8. 交替打印两个数组

	package test;
	//交替打印两个数组
	public class TwoArr {
	int []arr1 = new int[]{1,3,5,7,9};
	int []arr2 = new int[]{2,4,6,8,10};
	public static boolean flag = false;
	public synchronized void printArr1(){
	for(int i=0;i<arr1.length;i++){
	while (flag){
	try {
	this.wait();
	} catch (InterruptedException e) {
	e.printStackTrace();
	}
	}
	System.out.println(arr1[i]);
	flag=!flag;
	notifyAll();
	}
	}
	public synchronized void printArr2(){
	for(int i=0;i<arr2.length;i++){
	while (!flag){
	try {
	this.wait();
	} catch (InterruptedException e) {
	e.printStackTrace();
	}
	}
	System.out.println(arr2[i]);
	flag=!flag;
	notifyAll();
	}
	}
	public static void main(String[] args) {
	TwoArr twoArr = new TwoArr();
	new Thread(()->twoArr.printArr1()).start();
	new Thread(()->twoArr.printArr2()).start();
	}
	}

   

9. 猴子抢桃子多线程（多窗口变形）

	//3只猴子抢着分100只桃子，每只抢到的猴子分去剩余桃子的一半。使用多线程模拟这一过程
	public class Monet {
	static class Peach{
	int value = 100;
	int temp = 0;
	public void get(){
	while (true){
	temp = value;
	synchronized (this){
	if(temp>0){
	if(temp%2==0){
	value = temp/2;
	}else{
	value = temp-(temp+1)/2;
	}
	System.out.println(Thread.currentThread().getName() + " 抢到了 " + (temp - value) + " 个桃子，还剩 "+
	value);
	try {
	Thread.sleep(500);
	} catch (InterruptedException e) {
	e.printStackTrace();
	}
	}else{
	System.out.println("桃子已经无了");
	break;
	}
	}
	}
	}
	}
	static class Monkey extends Thread{
	Peach peach;
	public Monkey(String name, Peach peach) {
	super(name);
	this.peach = peach;
	}
	@Override
	public void run() {
	super.run();
	peach.get();
	}
	}
	public static void main(String[] args) {
	Peach peach = new Peach();
	Monkey m1 = new Monkey("猴子1", peach);
	Monkey m2 = new Monkey("猴子2", peach);
	Monkey m3 = new Monkey("猴子3", peach);
	m1.start();
	m2.start();
	m3.start();
	}
	}

   

10. 总结

首先做这个题目，就俩种形式，一种是顺序打印，就只有学会 wait (),notify (),notifyAll (), 然后，你要等线程做完了，才获取下一步则使用 Join（），然后关于多窗口卖票和猴子抢桃子，实际上是一种题目的变形，我只要保证 ticket 和 peach 是同一个对象，我就可以操作他们，以及怎么保持数据一致性，就使用 synchronized 解决

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