线程安全问题，我下面的想法是交替执行sout（i），为什么报错

Question

public class CommunicationTest {
    public static void main(String[] args) {

        Number n1 = new Number();
        n1.setName("线程1");
        n1.start();
        Number n2 = new Number();
        n2.setName("线程2");
        n2.start();
    }
}
class Number extends Thread{
    private  static  int i ;
    @Override
    public void run() {
            for(;;) {
                synchronized (Number.class) {
                    notify();
                if(i <= 100) {
                try {
                    Thread.sleep(500);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                    System.out.println(getName() + ":" + i);
                    i++;
                    try {
                        wait();
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                }else{
                    break;
                }
            }
        }
    }
}

Answer 1

(1)静态变量 i 没有初始化啊。
private static int i =0;
(2)notify()和wait()的使用不对。
修改后运行结果：

Number类修改如下：

public class Number extends Thread{
    private  static  int i =0;

    @Override
    public void run() {
        for(;;) {
            synchronized (Number.class) {
                Number.class.notify();//注意这里加上加锁的对象
                
                if(i <= 100) {
                    try {
                        Thread.sleep(1);
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                    System.out.println(getName() + ":" + i);
                    i++;

                    try {
                        Number.class.wait();//注意这里加上加锁的对象
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                }else{
                    break;
                }
            }
        }
    }
}