JAVA异步调用再度改进

NEW A（MANY）-》SEND MESSAGE(UNIQUE)
NEW B（MANY）-》SET MESSAGE(UNIQUE)

ONLY ONE C（IF A GET MESSAGE A SENT，C remove MESSAGE FROM HASHMAP）

public class AClient2C2 {

	private final long timeout = 10000;

	public void test(String taskId) {
		long count = 0;
		while (CData2C2.getInstance().isHasTask(taskId) && count < timeout) {

			try {
				System.out.println("wait+"+count);
				Thread.sleep(1000);
				count += 1000;
			} catch (InterruptedException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
		}
		if(count>=timeout) {
			System.out.println("超时");
		}

	}

	public static void main(String[] args){
		//B的方法会改变CData2C2的值
		CData2C2.getInstance().regesitTaskId("aaaa");
		AClient2C2 ac = new AClient2C2();
		ac.test("aaaa");
		//A的方法获取CData2C2的值会把对象从HASHMAP里删除
		CData2C2.getInstance().removeTaskId("aaaa");
	}
}

import java.util.HashMap;
import java.util.Map;

public class BService2C2 {

}

import java.util.HashMap;
import java.util.Map;

public class CData2C2 {

	private static CData2C2 instance;

	private static Map<String,String> map = new HashMap<String,String>();

    private CData2C2() {

    }

    public static CData2C2 getInstance(){
    	if(instance==null){
    		instance = new CData2C2();
    	}
    	return instance;
    }

    public void regesitTaskId(String taskId){
    	map.put(taskId, taskId);
    }

    public void removeTaskId(String taskId){
    	map.remove(taskId);
    }

    public boolean isHasTask(String taskId){
    	if(null!=map.get(taskId)){
    		return Boolean.TRUE;
    	}
    	return Boolean.FALSE;
    }

}