应用中双击返回键退出程序

1、重写onKeyDown();

 @Overridepublic boolean onKeyDown(int keyCode, KeyEvent event) {    if (keyCode == KeyEvent.KEYCODE_BACK) {        exit();        return false;    }    return super.onKeyDown(keyCode, event);}

2、exit()实现：

 //2秒内连续点击 public void exit() {    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);    }}