两个线程交替打印0~100的奇偶数？

这个问题是一个常见的多线程同步问题，可以使用Java中的wait和notify方法，或者更现代的Semaphore，CyclicBarrier，CountDownLatch，Lock和Condition等同步工具来解决。这里我会给出使用wait和notify的示例，以及使用Semaphore的示例。

使用wait和notify的示例

public class OddEvenPrinter {
    private int number = 1;
    private final int MAX_NUMBER = 100;
    private final Object oddLock = new Object();
    private final Object evenLock = new Object();

    public void printOdd() {
        while (number <= MAX_NUMBER) {
            synchronized (oddLock) {
                while (number % 2 == 0) {
                    try {
                        oddLock.wait();
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                }
                System.out.println(Thread.currentThread().getName() + ": " + number);
                number++;
                evenLock.notify();
            }
        }
    }

    public void printEven() {
        while (number <= MAX_NUMBER) {
            synchronized (evenLock) {
                while (number % 2 != 0) {
                    try {
                        evenLock.wait();
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                }
                System.out.println(Thread.currentThread().getName() + ": " + number);
                number++;
                oddLock.notify();
            }
        }
    }

    public static void main(String[] args) {
        OddEvenPrinter printer = new OddEvenPrinter();
        Thread oddThread = new Thread(() -> printer.printOdd(), "OddThread");
        Thread evenThread = new Thread(() -> printer.printEven(), "EvenThread");
        oddThread.start();
        evenThread.start();
    }
}

在这个示例中，我们创建了两个锁对象oddLock和evenLock，分别用于控制奇数和偶数的打印。printOdd方法只打印奇数，而printEven方法只打印偶数。当number为奇数时，printOdd方法会打印它并通知printEven方法可以继续执行；当number为偶数时，情况相反。我们使用wait方法让线程等待，直到条件满足（即number变为期望的奇偶性），然后使用notify方法唤醒等待的线程。

使用Semaphore的示例

import java.util.concurrent.Semaphore;

public class OddEvenPrinterWithSemaphore {
    private int number = 1;
    private final int MAX_NUMBER = 100;
    private final Semaphore oddSemaphore = new Semaphore(1);
    private final Semaphore evenSemaphore = new Semaphore(0);

    public void printOdd() throws InterruptedException {
        for (; number <= MAX_NUMBER; number += 2) {
            oddSemaphore.acquire();
            System.out.println(Thread.currentThread().getName() + ": " + number);
            evenSemaphore.release();
        }
    }

    public void printEven() throws InterruptedException {
        for (int i = 2; i <= MAX_NUMBER; i += 2) {
            evenSemaphore.acquire();
            System.out.println(Thread.currentThread().getName() + ": " + i);
            oddSemaphore.release();
        }
    }

    public static void main(String[] args) {
        OddEvenPrinterWithSemaphore printer = new OddEvenPrinterWithSemaphore();
        Thread oddThread = new Thread(() -> {
            try {
                printer.printOdd();
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }, "OddThread");
        Thread evenThread = new Thread(() -> {
            try {
                printer.printEven();
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }, "EvenThread");
        oddThread.start();
        evenThread.start();
    }
}

在这个示例中，我们使用两个Semaphore来控制奇数和偶数的打印。oddSemaphore初始时有一个许可，允许打印第一个奇数；evenSemaphore初始时没有许可，因此打印偶数的线程会立即阻塞。当打印奇数的线程打印完一个奇数并释放evenSemaphore的一个许可