【问题的描述：
启动3个线程打印递增的数字, 线程1先打印1,2,3,4,5, 然后是线程2打印6,7,8,9,10, 然后是线程3打印11,12,13,14,15. 接着再由线程1打印16,17,18,19,20....以此类推, 直到打印到75. 程序的输出结果应该为:
线程1: 1
线程1: 2
线程1: 3
线程1: 4
线程1: 5
线程2: 6
线程2: 7
线程2: 8
线程2: 9
线程2: 10
...
线程3: 71
线程3: 72
线程3: 73
线程3: 74
线程3: 75
【代码实现
import java.util.ArrayList;
import java.util.List;
/**
* Created by ibm on 2017/8/8.
*/
public class ClassicTest {
//使用原始的synchornized object.wait() object.notify()
public static void main(String[] args) {
//定义线程组
List<MyThread> threadGroups = new ArrayList<>();
Counter counter = new Counter();
MyThread t1 = new MyThread(1,"一",counter,threadGroups);
MyThread t2 = new MyThread(2,"二",counter,threadGroups);
MyThread t3 = new MyThread(2,"三",counter,threadGroups);
threadGroups.add(t1);
threadGroups.add(t2);
threadGroups.add(t3);
new Thread(t1).start();
new Thread(t2).start();
new Thread(t3).start();
}
}
class MyThread implements Runnable{
//线程运行状态 1马上执行，2阻塞
public int status;
//线程名字
public String name;
//计数器
public Counter counter;
//线程组
public List<MyThread> threads = new ArrayList<>();
public MyThread(int status,String name,Counter counter,List<MyThread> threads){
this.status = status;
this.name = name;
this.counter = counter;
this.threads = threads;
}
@Override
public void run() {
System.out.println(name + " GET " + status);
synchronized (counter){
while (!counter.isEnd()){
//判断是否该自己执行，切记使用while,因为如果在循环的等待过程中status有所变化，这里需要再次判断
while (status != 1){
try {
counter.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
for(int i = 0;i < 5 ;i ++){
System.out.println(name + ":" + counter.get());
counter.increase();
}
//状态进入阻塞状态，并设置下一个可以运行的线程
status = 2;
setNext();
counter.notifyAll();
System.out.println("----");
}
}
}
void setNext(){
//当前线程在线程组的索引
int index = 0;
for(index = 0;index < threads.size();index++){
if(name.equals(threads.get(index).name)){
break;
}
}
if(index == (threads.size() - 1)){
threads.get(0).status = 1;
}else {
threads.get(index + 1).status = 1;
}
}
}
class Counter{
int num = 1;
int end = 75;
public int get(){
return num;
}
public void increase(){
if(isEnd()){
System.out.println("num超过限制");
return;
}
num++;
}
public boolean isEnd(){
if(num >= end){
return true;
}
return false;
}
}
【后面两种解法来源网络：
1.使用synchronized关键字：
public class ClassicTest1 {
// n为即将打印的数字
private static int n = 1;
// state=1表示将由线程1打印数字, state=2表示将由线程2打印数字, state=3表示将由线程3打印数字
private static int state = 1;
public static void main(String[] args) {
final ClassicTest1 pn = new ClassicTest1();
new Thread(new Runnable() {
public void run() {
// 3个线程打印75个数字, 单个线程每次打印5个连续数字, 因此每个线程只需执行5次打印任务. 3*5*5=75
for (int i = 0; i < 5; i++) {
// 3个线程都使用pn对象做锁, 以保证每个交替期间只有一个线程在打印
synchronized (pn) {
// 如果state!=1, 说明此时尚未轮到线程1打印, 线程1将调用pn的wait()方法, 直到下次被唤醒
while (state != 1)
try {
pn.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
// 当state=1时, 轮到线程1打印5次数字
for (int j = 0; j < 5; j++) {
// 打印一次后n自增
System.out.println(Thread.currentThread().getName()
+ ": " + n++);
}
System.out.println();
// 线程1打印完成后, 将state赋值为2, 表示接下来将轮到线程2打印
state = 2;
// notifyAll()方法唤醒在pn上wait的线程2和线程3, 同时线程1将退出同步代码块, 释放pn锁.
// 因此3个线程将再次竞争pn锁
// 假如线程1或线程3竞争到资源, 由于state不为1或3, 线程1或线程3将很快再次wait, 释放出刚到手的pn锁.
// 只有线程2可以通过state判定, 所以线程2一定是执行下次打印任务的线程.
// 对于线程2来说, 获得锁的道路也许是曲折的, 但前途一定是光明的.
pn.notifyAll();
}
}
}
}, "线程1").start();
new Thread(new Runnable() {
public void run() {
for (int i = 0; i < 5; i++) {
synchronized (pn) {
while (state != 2)
try {
pn.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
for (int j = 0; j < 5; j++) {
System.out.println(Thread.currentThread().getName()
+ ": " + n++);
}
System.out.println();
state = 3;
pn.notifyAll();
}
}
}
}, "线程2").start();
new Thread(new Runnable() {
public void run() {
for (int i = 0; i < 5; i++) {
synchronized (pn) {
while (state != 3)
try {
pn.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
for (int j = 0; j < 5; j++) {
System.out.println(Thread.currentThread().getName()
+ ": " + n++);
}
System.out.println();
state = 1;
pn.notifyAll();
}
}
}
}, "线程3").start();
}
}2.使用condition与lock
public class ClassicTest2 implements Runnable {
private int state = 1;
private int n = 1;
// 使用lock做锁
private ReentrantLock lock = new ReentrantLock();
// 获得lock锁的3个分支条件
private Condition c1 = lock.newCondition();
private Condition c2 = lock.newCondition();
private Condition c3 = lock.newCondition();
@Override
public void run() {
new Thread(new Runnable() {
public void run() {
for (int i = 0; i < 5; i++) {
try {
// 线程1获得lock锁后, 其他线程将无法进入需要lock锁的代码块.
// 在lock.lock()和lock.unlock()之间的代码相当于使用了synchronized(lock){}
lock.lock();
while (state != 1)
try {
// 线程1竞争到了lock, 但是发现state不为1, 说明此时还未轮到线程1打印.
// 因此线程1将在c1上wait
// 与解法一不同的是, 三个线程并非在同一个对象上wait, 也不由同一个对象唤醒
c1.await();
} catch (InterruptedException e) {
e.printStackTrace();
}
// 如果线程1竞争到了lock, 也通过了state判定, 将执行打印任务
for (int j = 0; j < 5; j++) {
System.out.println(Thread.currentThread().getName()
+ ": " + n++);
}
System.out.println();
// 打印完成后将state赋值为2, 表示下一次的打印任务将由线程2执行
state = 2;
// 唤醒在c2分支上wait的线程2
c2.signal();
} finally {
// 打印任务执行完成后需要确保锁被释放, 因此将释放锁的代码放在finally中
lock.unlock();
}
}
}
}, "线程1").start();
new Thread(new Runnable() {
public void run() {
for (int i = 0; i < 5; i++) {
try {
lock.lock();
while (state != 2)
try {
c2.await();
} catch (InterruptedException e) {
e.printStackTrace();
}
for (int j = 0; j < 5; j++) {
System.out.println(Thread.currentThread().getName()
+ ": " + n++);
}
System.out.println();
state = 3;
c3.signal();
} finally {
lock.unlock();
}
}
}
}, "线程2").start();
new Thread(new Runnable() {
public void run() {
for (int i = 0; i < 5; i++) {
try {
lock.lock();
while (state != 3)
try {
c3.await();
} catch (InterruptedException e) {
e.printStackTrace();
}
for (int j = 0; j < 5; j++) {
System.out.println(Thread.currentThread().getName()
+ ": " + n++);
}
System.out.println();
state = 1;
c1.signal();
} finally {
lock.unlock();
}
}
}
}, "线程3").start();
}
public static void main(String[] args) {
new ClassicTest2().run();
}
}
