FD DS 面经

select distinct contact_a, count(distinct contact_b) from (
select sender_id as contact_a, receiver_id as contact_b from (select distinct sender_id, receiver_id from #message)a
union all
select receiver_id as contact_a, sender_id as contact_b from (select distinct sender_id, receiver_id from #message)a )c
group by contact_a

select  total, count(id) from
(select  id ,sum(total) from ((select  a.sender_id 'id',count(*) 'total' from message as a group by a.sender_id)
union
(select b.receiver_id 'id',count(*) 'total' from message as b group by
b.sender_id ))
group by id)
group by total

raichi 发表于 2017-11-29 11:11
我觉得这么写不对，题目问的是“某一天收到或者发过的信息的unique user 数目”
如果有一个用户收到信息 ...

感觉union all可能是最好的答案？

entropy2017 发表于 2018-3-25 09:11
感觉union all可能是最好的答案？

select d.total, count(d.id)
from (select c.id, count(c.id) as total
         from (select sender_id as id
       from (select distinct sender_id, receiver_id from message) a .鍦帮紝鐙
       union all
       select receiver_id as id
                  from (select distinct sender_id, receiver_id from message) b) c
          group by c.id) d
group by d.total

thanks for sharing!

楼主觉得新算法是好是坏呢？我自己稍微想一下觉得还是好的，因为acceptance rate虽然没变但是count应该增加了，所以算法没有太大问题？

看了几个FB DS面经下来，貌似FB也是拿DS当DA来面。弱问一句lz，是PhD还是小M？