再按一次退出程序

 private long exitTime = 0
 public boolean onKeyDown(int keyCode, KeyEvent event) {
  if(keyCode == KeyEvent.KEYCODE_BACK && event.getAction() == KeyEvent.ACTION_DOWN){  
      if((System.currentTimeMillis()-exitTime) > 2000){ 
       Log.d("log", "----------------------->"+String.valueOf(System.currentTimeMillis()));
          Toast.makeText(getApplicationContext(), "再按一次退出程序", Toast.LENGTH_SHORT).show();                               
          exitTime = System.currentTimeMillis();  
      }  
      else{  
         finish();  
         System.exit(0);  
      }  
      return true;  
     }
  return super.onKeyDown(keyCode, event);
 }

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原文地址:https://www.cnblogs.com/zhiqiangsimida/p/2774700.html