Java 多线程练习

题目：某公司组织年会,会议入场时有两个入口,在入场时每位员工都能获取一张双色球彩票,假设公司有100个员工,利用多线程模拟年会入场过程,

并分别统计每个入口入场的人数,以及每个员工拿到的彩票的号码。线程运行后打印格式如下：

编号为: 2 的员工 从后门 入场! 拿到的双色球彩票号码是: [17, 24, 29, 30, 31, 32, 07]

编号为: 1 的员工 从后门 入场! 拿到的双色球彩票号码是: [06, 11, 14, 22, 29, 32, 15]

//.....

从后门入场的员工总共: 13 位员工

从前门入场的员工总共: 87 位员工

 

分析：两个入口对应两个线程，获取彩票的方法覆写runnable接口的run方法实现，我把这个实现了runnable接口的类称作paper类，里面用id字段记录员工编号，并提供get，set方法。设置第3个线程来作为随机控制100个员工进出哪个入口，因为是员工进入了入口，入口才相应彩票方法，所以我用了等待唤醒机制(这里要注意，这个等待唤醒机制必须要在同步块中，还要用同一把锁。)

 

代码：

class paper implements runnable {

    boolean bflag = false;

    private int id;

    public void setid(int id) {

        this.id = id;
    }

    public int getid() {
        return this.id;
    }

    public string position;

    private list<string> listposcount = new arraylist<string>();

    private hashmap<integer, int[]> dic = new hashmap<integer, int[]>();

    private list<integer> mylist = new arraylist<integer>();

    private int[] getnumbers(int size) {

        int numbers[] = new int[size];

        for (int i = 0; i < size;) {

            boolean flag = false;

            numbers[i] = (int) (math.random() * 100);

            for (int j = 0; j < i; j++) {

                if (numbers[j] == numbers[i]) {

                    flag = true;

                    break;

                }
            }

            if (flag == true) {

                continue;
            }

            i++;
        }
        return numbers;
    }

    @override
    public void run() {
        synchronized (this) {
            while (mylist.size() < 100) {

                while (!bflag) {

                    try {

                        this.wait();

                    } catch (interruptedexception e) {

                        e.printstacktrace();
                    }
                }        
            
                this.bflag = false;
                
                if (!mylist.contains(id)) {

                    mylist.add(id);
                    if (position == "qian门" || position == "后门")
                        listposcount.add(position);

                    int nums[] = getnumbers(7);

                    dic.put(id, nums);

                    system.out
                            .println("编号为: " + id + " 的员工 从" + position + " 入场! 拿到的双色球彩票号码是:" + arrays.tostring(nums));

                } else {

                    
                }
                this.notifyall();
            }

            int a = 0, b = 0;
            for (string str : listposcount) {

                if (str == "后门")
                    a++;
                if (str == "qian门")
                    b++;

            }
            system.out.println("从后门入场的员工总共:" + a + " 位员工");
            system.out.println("从后门入场的员工总共:" + b + " 位员工");

        }
    }
}

//--------------------------------------------------------------------------------

//测试main函数

public class threadlearn2 {

	public static void main(string[] args) {

		paper p = new paper();

		thread th1 = new thread(p);
		thread th2 = new thread(p);

		runnable run1 = new runnable() {

			@override
			public void run() {
				synchronized (p) {
					
				
					list<integer> list = new arraylist<integer>();

					while (list.size() < 100) {
						int i = (int) (math.random() * 100);

						if (list.indexof(i) == -1) {

							p.setid(i);

							list.add(i);

							p.bflag = true;

							int num = (int) (math.random() * 100) % 2;

							if (num == 1) {

								p.position = "后门";
							} else {
								p.position = "qian门";

							}
							p.notifyall();
							
							while (p.bflag) {

								try {

									p.wait();

								} catch (interruptedexception e) {

									e.printstacktrace();
								}
							}

						}

					}

				}
			}

		};

		thread th3 = new thread(run1);

		th1.start();
		th2.start();
		th3.start();

	}

}