线程题：交替打印1-100

这里演示两个线程，一个打印奇数，一个打印偶数

方式一：synchronized + FixedThreadPool

public class example {private static int count = 1;private static final Object lock = new Object();public static void main(String[] args) {ThreadFactory factory = new ThreadFactory() {AtomicInteger ai = new AtomicInteger(1);@Overridepublic Thread newThread(@NotNull Runnable r) {Thread thread = new Thread(r, ai.getAndIncrement() + ": ");return thread;}};ExecutorService executorService = Executors.newFixedThreadPool(2, factory);executorService.submit(example::printOdd);executorService.submit(example::printEven);}private static void printOdd() {while (count <= 100) {synchronized (lock) {if(count % 2 != 0 && count <= 100) {System.out.println(Thread.currentThread().getName() + ": " + count);count++;}}}}private static void printEven() {while (count <= 100) {synchronized (lock) {if(count % 2 == 0 && count <= 100) {System.out.println(Thread.currentThread().getName() + ": " + count);count++;}}}}
}

方法二：ReentrantLock + Condition

public class example2 {static ReentrantLock lock = new ReentrantLock();static int count = 1;public static void main(String[] args) {// 两个 conditionCondition odd = lock.newCondition();Condition even = lock.newCondition();new Thread(() -> {while(count <= 100) {lock.lock();try {try {if (count % 2 != 0) {System.out.println(Thread.currentThread().getName() + ": " + count);count++;}even.signal();odd.await();} catch (InterruptedException e) {throw new RuntimeException(e);}}finally {lock.unlock();}}}, "odd").start();new Thread(() -> {while(count <= 100) {lock.lock();try {try {if (count % 2 == 0) {System.out.println(Thread.currentThread().getName() + ": " + count);count++;}odd.signal();even.await();} catch (InterruptedException e) {throw new RuntimeException(e);}}finally {lock.unlock();}}}, "even").start();}
}