android 再按一次后退键退出应用程序

方法一：
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;
    }

方法2：

public class BackActivity extends Activity {

        private static final String TAG = "BackActivity";
        private long mLastBackTime = 0;
        private long TIME_DIFF = 2 * 1000;

        @Override
        public void onCreate(Bundle savedInstanceState) {
                super.onCreate(savedInstanceState);
                setContentView(R.layout.main);
        }

        @Override
        public boolean onKeyDown(int keyCode, KeyEvent event) {
                if (keyCode == KeyEvent.KEYCODE_BACK) {
                        long now = new Date().getTime();
                        if (now - mLastBackTime < TIME_DIFF) {
                                Log.d(TAG, "back, finish");
                                return super.onKeyDown(keyCode, event);
                        } else {
                                Log.d(TAG, "after 2s, toast");
                                mLastBackTime = now;
                                Toast.makeText(this, "再点一次将推出", 2000).show();
                        }
                        return true;
                }
                return super.onKeyDown(keyCode, event);
        }
}