对于电梯内发出的请求，方法不变，对于电梯外发出的任务请求，可以用两个队列分别保存向上和向下的请求，在各电梯处理电梯内发出的请求的时候，每到达一层，就检查上述的两个队列，看是否有匹配项，若有则停留3秒，从队列中去除任务，若有电梯没有电梯内发出的请求可处理而停滞，则只要上述的两个队列不空，就赶去处理，这种方法从代码方面来说相对简单，但这样做的话任务的完成具有不可预测性，也会造成一个请求，多个电梯响应的问题，不知道老师怎么想的？(For the elevator the request method for the same, a lift task request, you can use two queue stored up and down the request, when the elevator elevator processing request, each reached a two layer, check the queue, see if there is a match, if there is a stop for 3 seconds, remove the task from the queue, if there is no elevator elevator in the request processing can be stopped, as long as the two queue is not empty, the rush to the processing, this method from the aspects of the code is relatively simple, but do this task is unpredictable. Will cause a request, a plurality of lift response problems, teachers do not know how to think?)