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编写程序Java，创建三个新线程

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编写一个程序Java，该程序创建三个新线程（已经存在的主线程除外） 并使它们同步，以使每个线程依次显示其线程ID进行5次迭代。 程序的输出应如下所示：

线程1-迭代编号。 1

线程2-迭代编号1

线程3-迭代编号1

线程1-迭代编号。 2

线程2-迭代编号2

线程3-迭代编号2

线程1-迭代编号。 3

线程2-迭代编号3

线程3-迭代编号3

线程1-迭代编号。 4

线程2-迭代编号4

线程3-迭代编号4

线程1-迭代编号。 5

线程2-迭代编号5

线程3-迭代编号5

代码： 

package ZTest;
public class PrintSequenceRunnable2 implements Runnable{

    public int PRINT_NUMberS_UPTO=15;
    static int  number=1;
    int remainder;
    static Object lock=new Object();

    PrintSequenceRunnable2(int remainder)
    {
        this.remainder=remainder;
    }

    public void run() {
        while (number < PRINT_NUMberS_UPTO-1) {
            synchronized (lock) {
                while (number % 3 != remainder) { 
                    try {
                        lock.wait();
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                }
                System.out.println(Thread.currentThread().getName() + "- iteration no. " + number);
                number++;
                lock.notifyAll();
            }
        }
    }
    public static void main(String[] args) {

        PrintSequenceRunnable2 runnable1=new PrintSequenceRunnable2(1);
        PrintSequenceRunnable2 runnable2=new PrintSequenceRunnable2(2);
        PrintSequenceRunnable2 runnable3=new PrintSequenceRunnable2(0);

            Thread t1=new Thread(runnable1,"Thread1");
            Thread t2=new Thread(runnable2,"Thread2");
            Thread t3=new Thread(runnable3,"Thread3");

            t1.start();
            t2.start();
            t3.start();   
        }
    }

我的输出： 

Thread1- iteration no. 1
Thread2- iteration no. 2
Thread3- iteration no. 3
Thread1- iteration no. 4
Thread2- iteration no. 5
Thread3- iteration no. 6
Thread1- iteration no. 7
Thread2- iteration no. 8
Thread3- iteration no. 9
Thread1- iteration no. 10
Thread2- iteration no. 11
Thread3- iteration no. 12
Thread1- iteration no. 13
Thread2- iteration no. 14
Thread3- iteration no. 15

预期输出： 

Thread 1 - iteration no. 1

Thread 2 - iteration no. 1

Thread 3 - iteration no. 1

Thread 1 - iteration no. 2

Thread 2 - iteration no. 2

Thread 3 - iteration no. 2

Thread 1 - iteration no. 3

Thread 2 - iteration no. 3

Thread 3 - iteration no. 3

Thread 1 - iteration no. 4

Thread 2 - iteration no. 4

Thread 3 - iteration no. 4

Thread 1 - iteration no. 5

Thread 2 - iteration no. 5

Thread 3 - iteration no. 5
hail1999 回答：编写程序Java，创建三个新线程

我发现这是一个奇怪的问题，因为您基本上是在实现线程，同时也完全取消了对线程的使用。

但这是我的建议：

当前正在打印以下内容

System.out.println(Thread.currentThread().getName() + "- iteration no. " + number);

使数字遵循所需方案的最简单方法是将计数器除以线程数，然后对结果调用math.floor。像这样：

int iteration = Math.floor(number / 3.0) + 1; //3.0 is the amount of threads

这将保存您要的结果。请记住，由于计数器从1开始，因此您可能必须从number中减去1才能使其正确对齐。

,

这是执行此操作的一种粗略方法。整个想法是，一个线程解锁另一个线程的执行。您可以使用循环屏障来做同样的事情。

databaseReference.addValueEventListener(new ValueEventListener() {
            @Override
            public void onDataChange(@NonNull DataSnapshot dataSnapshot) {

                for (final DataSnapshot parentSnap : dataSnapshot.getChildren()) {

                    if (parentSnap.exists()) {
                        progressDialog.dismiss();
                        SurveysListModel model = parentSnap.getValue(SurveysListModel.class);
                        list.add(model);

                        adapter = new SurveysListAdapter(list,UserAllSurveys.this,new SurveysListAdapter.HandleClickListener() {
                            @Override
                            public void onItemClicked(int position,SurveysListModel surveysListModel) {
                                String typ = surveysListModel.getSurvey_name();
                                Toast.makeText(UserAllSurveys.this,"CLicked" + position,Toast.LENGTH_SHORT).show();
                            }
                        });
                        recyclerView.setAdapter(adapter);
                    }
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