更快速地检查特定端口的服务器？

Question

我正在尝试使用java / Android检查网络中特定端口上运行的所有服务器。我正在使用的方法目前需要5-10分钟才能做到这一点。有更快的方法吗？

我是socket编程的新手，所以任何帮助都将受到高度赞赏。

public void findservers()//throws UnknownHostException, SocketException
    {


        Enumeration e = null;
        try {
            e = NetworkInterface.getNetworkInterfaces();
        } catch (SocketException e1) {
            e1.printStackTrace();
        }
        //Data dt = new Data();
        //Client cl = new Client();
        //dt.setClientObject(cl);
        while(e.hasMoreElements())
        {
            NetworkInterface n = (NetworkInterface) e.nextElement();
            Enumeration ee = n.getInetAddresses();
            while (ee.hasMoreElements())
            {
                InetAddress i = (InetAddress) ee.nextElement();
                String ip = i.getHostAddress();
                System.out.println(ip);

                String sip = ip.substring(0, ip.indexOf('.',ip.indexOf('.',ip.indexOf('.')+1) + 1) + 1);
                System.out.println(sip);
                try {
                    for(int it=1;it<=255;it++)
                    {
                        String sip1 = sip+it;
                        System.out.println(sip1);
                        //InetAddress.getByName(sip1).isReachable(100);
                        //System.out.println(sip1+" is online");
                        findserver(sip1);
                        //cl.ask(sip);

                    }
                } catch (Exception e1) {
                    System.out.println(sip);
                    System.out.println("hello");
                }
            }
        }

    }
 public void findserver(String ip)
    {
       // String ip = "192.168.43.41";

        try
        {
            InetAddress ping = InetAddress.getByName(ip);
            Socket s = new Socket(ping, 50005);
            System.out.println("Server found on IP: " + ping.getCanonicalHostName());
            s.close();
        } catch (IOException e) {
            System.out.println("Nothing much");
            System.out.println(e);
        }
    }