2秒内双击返回退出应用

@Override
    public boolean onKeyDown(int keyCode, KeyEvent event) {

        if (keyCode == KeyEvent.KEYCODE_BACK) {

            if (System.currentTimeMillis() - mExitTime > 2000) {

                Toast.makeText(getApplicationContext(), "再按一次返回键退出",
                        Toast.LENGTH_SHORT).show();
                mExitTime = System.currentTimeMillis();
            } else {
                finish();
                ExitApplication.getInstance().exit();
            }
            return true;
        }
        return super.onKeyDown(keyCode, event);
    }