JAVA 兩個線程，一個線程打印1-10，另一個線程打印1-20，線程互不搶佔，輪流打印20遍 用信號量如何實現?

import java.util.Random;
import java.util.concurrent.Semaphore;

public class Test {

    public static void main(String[] args){
        Semaphore semaphore = new Semaphore(1);
        final boolean[] shouldThread1 = {true};
        int cycleTime = 20;
        new Thread(new Runnable() {
            @Override
            public void run() {
                Random random = new Random();
                for(int i=1;i<=cycleTime;i++){
                    while (true){
                        if(shouldThread1[0]){
                            try {
                                semaphore.acquire();
                            break;
                            } catch (InterruptedException e) {
                                e.printStackTrace();
                            }
                        }else{
                            try {
                                semaphore.acquire();
                            } catch (InterruptedException e) {
                                e.printStackTrace();
                            }finally {
                                semaphore.release();
                            }
                        }
                    }
                    for(int j=1;j<=10;j++){
                        System.out.println( "loop" + i + " 線程1:  " + j );
                        try {
                            Thread.sleep( random.nextInt( 20 ) );
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                    System.out.println();
                    shouldThread1[0] =false;
                    semaphore.release();
                }
            }
        }).start();

        new Thread(new Runnable() {
            @Override
            public void run() {
                Random random = new Random();
                for(int i=1;i<=cycleTime;i++){
                    while (true){
                        if( !shouldThread1[0] ){
                            try {
                                semaphore.acquire();
                            break;
                            } catch (InterruptedException e) {
                                e.printStackTrace();
                            }
                        }else{
                            try {
                                semaphore.acquire();
                            } catch (InterruptedException e) {
                                e.printStackTrace();
                            }finally {
                                semaphore.release();
                            }
                        }
                    }
                    for(int j=1;j<=20;j++){
                        System.out.println( "                             loop" + i + " 線程2:  "+ j );
                        try {
                            Thread.sleep( random.nextInt( 40 ) );
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                    System.out.println();
                    shouldThread1[0]=true;
                    semaphore.release();
                }
            }
        }).start();
    }
}