请看我对这个问题的评论，主要的一个是关于个人偏好的，整个想法毫无意义。但是，目前尚不清楚，也许你已经解决了它。可以通过某种方式询问个人偏好来解决。每次我必须为我的门票预订固定座位时，我被问及我的偏好，有一些选项跟我一起。



现在，它看起来像你的算法，实际的时间复杂度是O（N），更糟糕的情况。实际上，你有N个座位，可以检查所有N，作为一张票的起点。它已经给你O（N）了。对于每个这样的座位，您需要在此点附近检查最多X * Y座位。但X * Y是一个常数参数。你没有测试y的所有值从0到X - 1，并且对Y来说都是相同的。这个常量带括号，它不会影响复杂性。这是因为常数因子不会改变函数的渐近行为。



请参阅：

http：/ /en.wikipedia.org/wiki/Computational_complexity_theory [ ^ ]，

http://en.wikipedia.org/wiki/Time_complexity [ ^ ]，

http://en.wikipedia.org/wiki/Big_O_notation [ ^ ]。

-SA

Please see my comments to the question, and the main one would be the first, about personal preference, which render the whole idea making little sense. However, it's not clear, maybe you already solved it. It can be solved by asking about personal preferences in some way. Every time I had to book fixed seats for my tickets, I was asked about my preferences, with some options followed with me.

Now, it looks like with your algorithm, the actual time complexity is O(N), in a worse case. Indeed, you have N seats, and can check all N, as a starting point for just one ticket. It gives you O(N) already. With each such seat, you need to check up at most X*Y seats around this point. But X*Y is a constant parameter. You don't test all values of y from 0 to X − 1, and the same for Y. This constant "carries out of brackets", it does not effect the complexity. This is because the constant factor does not change asymptotic behavior of a function.

Please see:
http://en.wikipedia.org/wiki/Computational_complexity_theory[^],
http://en.wikipedia.org/wiki/Time_complexity[^],
http://en.wikipedia.org/wiki/Big_O_notation[^].

—SA

添加谢尔盖所说的（在那个发现座位是O（N），但实际上并没有说什么是***的。



您可以为每个座位分配重量，更好的座位具有更高的值UE的。在搜索座位时，您将当前所选座位保持为最高值（y的总和（其中y是所需座位的数量），每个座位连续排座位）。如果您发现y座位值较高，请选择这些座位。



这可以归结为一个简单的排序算法，您可以从列表中删除占用的座位，然后根据索引进行排序。之后，您只需选择连续的前y座位。

To add to what Sergey said (in that finding the seats is O(N), but really doesn't say what ones are the "best".

You could assign weights to each seat, with "better" seats having higher values. While searching for seats you hold the currently selected seats as the highest value (sum of y (where y is the number of required seats) seats in a row for each guest). If you find y seats with a higher value, select those.

This then can come down to a simple sorting algorithm, where you remove occupied seats from the list and then sort based on index. After that you just select the top y seats that are sequential.