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惯例电面完发面经攒人品。1小时前的电面，意外的简单。先是5分钟－10分钟的自我介绍和聊project。然后上codepair写题，给一个mxn矩阵，每个点要么是建筑物要么是路，给你一个司机的位置，和一些乘客的位置，要求返回这个司机能

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补充内容 (2017-1-27 06:29):
去career fair找recruiter聊了一下，然后拿到了onsite

RobinSwift 发表于 2017-1-27 01:32
Why needs bfs or dfs here? Won't calculate the distance to driver and find the min directly with O(n ...

You only need to know the list of reachable riders. The output doesn't have to be sorted in distance.

yangluphil 发表于 2017-1-27 06:32
You only need to know the list of reachable riders. The output doesn't have to be sorted in distan ...

Thanks for the explanation! As long as there are paths to the potential riders, that rider can be the one that the drive can pick up. Very practical question.

RobinSwift 发表于 2017-1-27 01:32
Why needs bfs or dfs here? Won't calculate the distance to driver and find the min directly with O(n ...

You only need to know the list of reachable riders. The output doesn't have to be sorted by distance.

补充内容 (2017-1-27 06:49):
刚才卡了，重发了以上几条信息，求版主删除

请问楼主是什么时候海投的呀？

醉梦听雪 发表于 2017-1-22 17:15
请问楼主是什么时候海投的呀？

12/3 投的

Why needs bfs or dfs here? Won't calculate the distance to driver and find the min directly with O(n) where n is the number of potential passengers?

楼主，这个并机跑test case能细说一下吗？因为没遇到过。十分感谢

RobinSwift 发表于 2017-1-27 01:32
Why needs bfs or dfs here? Won't calculate the distance to driver and find the min directly with O(n ...

You only need to know the list of reachable riders. The output doesn't have to be sorted in distance.

david.fang 发表于 2017-1-27 01:43
楼主，这个并机跑test case能细说一下吗？因为没遇到过。十分感谢

就是自己写test case然后compile你的code并输出，不是口述test case的意思