双击返回键退出

1、监听返回键

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

2、逻辑判断

private 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);
}
}