[Android]2次返回键退出程序

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

貌似介个是最简单的一个了