题解 | #连续两次作答试卷的最大时间窗#

select uid,days_window,round(exam_cnt/days_off*days_window,2) as avg_exam_cnt
from(
select uid,count(start_time) as exam_cnt,
       datediff(max(start_time),min(start_time)) + 1 as days_off,
       max(datediff(next_start_time,start_time)) + 1 as days_window
from(
select *,lead(start_time)over(partition by uid order by start_time) as next_start_time
from exam_record
where year(start_time) = 2021) as t1
group by uid) as t2
where days_off > 1
order by days_window desc,avg_exam_cnt desc

先利用lead函数找到下一次的答题时间，然后按照题目要求去找最大时间间隔和答题的时间差，最后按照题目要求输出即可。