android 再按一次后退键退出应用程序

private static Boolean isExit = false;
    private static Boolean hasTask = false;
    Timer tExit = new Timer();
    TimerTask task = new TimerTask() {
         
        @Override
        public void run() {
            isExit = false;
            hasTask = true;
        }
    };
     
    @Override
    public boolean onKeyDown(int keyCode, KeyEvent event) {
        System.out.println("TabHost_Index.java onKeyDown");
        if (keyCode == KeyEvent.KEYCODE_BACK) {
            if(isExit == false ) {
                isExit = true;
                Toast.makeText(this, "再按一次后退键退出应用程序", Toast.LENGTH_SHORT).show();
                if(!hasTask) {
                    tExit.schedule(task, 2000);
                }
            } else {
                finish();
                System.exit(0);
            }
        }
        return false;
    }

简洁点的：

private long waitTime = 2000;  
	private long touchTime = 0; 
	@Override
	public boolean onKeyDown(int keyCode, KeyEvent event) {
		if (event.getAction() == KeyEvent.ACTION_DOWN && KeyEvent.KEYCODE_BACK == keyCode) {
			long currentTime = System.currentTimeMillis();
			if ((currentTime - touchTime) >= waitTime) {
				Toast.makeText(context, "再按一次退出程序", Toast.LENGTH_SHORT).show();
				touchTime = currentTime;
			} else {
				finish();
				System.exit(0);
			}
			return true;
		}
    	
		return super.onKeyDown(keyCode, event);
	}