思路1：记录上次点击的时间，与本次点击的时间比较，当两次时间间隔小于一定值时，退出，否则提示“再按一次退出程序”，同时更新上次点击时间

     private long firstTime = 0;     @Override  public boolean onKeyUp(int keyCode, KeyEvent event) {         // TODO Auto-generated method stub         switch(keyCode)         {         case KeyEvent.KEYCODE_BACK:              long secondTime = System.currentTimeMillis();                if (secondTime - firstTime > 2000) {                                         //如果两次按键时间间隔大于2秒，则不退出                   Toast.makeText(this, "再按一次退出程序", Toast.LENGTH_SHORT).show();                    firstTime = secondTime;//更新firstTime                   return true;                } else {                                                    //两次按键小于2秒时，退出应用              System.exit(0);               }              break;         }       return super.onKeyUp(keyCode, event);     }

思路2：开线程延时处理

private int mBackKeyPressedTimes = 0;            @Override          public void onBackPressed() {                  if (mBackKeyPressedTimes == 0) {                          Toast.makeText(this, "再按一次退出程序 ", Toast.LENGTH_SHORT).show();                          mBackKeyPressedTimes = 1;                          new Thread() {                                  @Override                                  public void run() {                                          try {                                                  Thread.sleep(2000);                                          } catch (InterruptedException e) {                                                  e.printStackTrace();                                          } finally {                                                  mBackKeyPressedTimes = 0;                                          }                                  }                          }.start();                          return;                        else{                                 this.activity.finish();                              }                  }                  super.onBackPressed();          }