报个Uber 的offer + 面经【一亩三分地论坛找工求职版】 - Powered by Discuz!


报个Uber 的offer + 面经【一亩三分地论坛找工求职版】 - Powered by Discuz!
1. 3个长度一样的array a1, a2, a3, 找出所有 A + B = C 的组合,A在a1里,B在a2,C在a3里;扩展到4个数组 a2, a2, a3, a4,找出A+B+C=D的组合。。 然后扩展到n各数组;做完了又给了一道题目,不难,忘了。
http://www.mitbbs.com/article_t/JobHunting/32945071.html
这题一个straightforward的方法(针对三个数组)就是先对每个数组排序,然后遍历C
中每一个元素,每个iteration用两个指针,一个指向A的头,一个指向B的尾,逐渐逼
近看能不能达到A+B=C。
这个复杂度是O(N^2)。

2. 一个系统里面,有user 在不断的login and logout, 现在给你几组user的(username, login_time, logout_time)的数据,打印出各个时间点系统有几个active user;follow-up,修改我的实现,让我的函数支持 类似JS的callback机制

第二题把所有时间排序 遇到login +1 logout -1, 一遍把所有要查看的时间点扫完 nlogn
优化的话,用两个hashtable存储某个时间点新上线和下线的人数,所有user过一遍,然后只对开始的时间点上线人数和结束的时间点下线人数加一。再过一遍所有的时间点,根据hashtable对上一个时间点update得到当前时间点人数,这样复杂度就是O(max(m, n))

3. 面我的是一个中国人神牛。因为我说我用c++,然后感觉他临时选了这道题给我。。 reader-writer problem. 用mutex 实现user-writer problem。 follow-up,问会有什么问题,其实会有starving 的情况发生,然后一起商量怎么处理
http://massivetechinterview.blogspot.com/2015/08/custom-readwritelock-and.html
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