title: onLeScan called twice when discovering bluetooth le devices
body: <p>Im trying to discover bluetooth low energy devices but I have a problem with onLeScan method. It is getting called twice and as a result all devices doubled. </p>
  
  <p>Method that i use to initiate scanning:</p>
  
  <pre><code>private void scanLeDevice(final boolean enable) {
          if (enable) {
              handler.postDelayed(new Runnable() {
                  @Override
                  public void run() {
                      bluetoothAdapter.stopLeScan(getLeScanCallback());
                      scanning = false;
                  }
              }, SCAN_PERIOD);
  
              scanning = true;
              bluetoothAdapter.startLeScan(getLeScanCallback());
          } else {
              bluetoothAdapter.stopLeScan(getLeScanCallback());
              scanning = false;
          }
      }
  </code></pre>
  
  <p>Method returning LeScanCallback:</p>
  
  <pre><code>private BluetoothAdapter.LeScanCallback getLeScanCallback(){
          BluetoothAdapter.LeScanCallback leScanCallback =
                  new BluetoothAdapter.LeScanCallback() {
                      @Override
                      public void onLeScan(final BluetoothDevice bluetoothDevice, int rssi
                              , byte[] scanRecord) {
                          runOnUiThread(new Runnable() {
                              @Override
                              public void run() {
                                  Log.d(TAG, "device found");
                                  Device device = new Device();
                                  device.setName(bluetoothDevice.getName());
                                  device.setAddress(bluetoothDevice.getAddress());
                                  devices.add(device);
                                  deviceListAdapter.notifyDataSetChanged();
                              }
                          });
                      }
                  };
  
          return leScanCallback;
      }
  </code></pre>