题解 | #牛客每个人最近的登录日期(四)#

解法：left join

select l1.date,count(distinct if(l2.date is null,l1.user_id,null)) 
from login l1
left join login l2
on  l1.user_id=l2.user_id and l1.date>l2.date
group by l1.date

易错解法：遗漏没有新登录用户的日期

select  fir_date,count(user_id)
from (
select user_id,min(date) as fir_date from login
group by user_id ) a 
group by fir_date