兩個線程交替打印1-100之間的奇偶數

實現一：synchronized關鍵字實現（缺點：效率低）

實現思路：讓兩個線程來爭奪同一把鎖，如果搶到鎖只有滿足自己的奇偶判定條件纔打印（注：&方式判斷奇偶更高效）

public class Demo04_兩個線程交替打印奇數_純synchronized {
    private static Object lock = new Object();
    private static int i = 1;
    public static void main(String[] args) throws InterruptedException {
        Thread t1 = new Thread(() -> {
            while (i < 100) {
                synchronized (lock) {
                    if ((i & 1) == 1) {
                        System.out.println(i++ + "是奇數");
                    }
                }
            }
        });
        Thread t2 = new Thread(() -> {
            while (i < 100) {
                synchronized (lock) {
                    if ((i & 1) == 0) {
                        System.out.println(i++ + "是偶數");
                    }
                }
            }
        });
        t1.start();;
        t2.start();
    }
}

實現二：wait/notify()（效率高）

思路：進行進程間的通信，奇線程打印完阻塞自己，偶線程打印完喚醒被阻塞的線程，同時將自己阻塞

public class Demo03_兩個線程交替打印奇偶數 {
    private static Object lock = new Object();
    private static int i = 1;
    public static void main(String[] args) {
        Thread t1 = new Thread(() -> {
            for (; i <= 100;) {
                synchronized (lock) {
                        System.out.println(i ++ + "：我是偶數");
                        try {
                            lock.notify();
                            if (i <= 100) {
                                lock.wait();
                            }
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                }
            }
        });
        Thread t2 = new Thread(() -> {
            for (; i <= 100;) {
                synchronized (lock) {
                        System.out.println(i ++ + "：我是奇數");
                        try {
                            if (i <= 100) {
                                lock.wait();
                            }
                            lock.notify();
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                }
            }
        });
        t2.start();
        t1.start();
    }