LeetCode（数据库）- 访问日期之间最大的空档期

题目链接：​​点击打开链接​​

题目大意：略。

解题思路：注意 LEAD / LAG 函数。

AC 代码

-- 解决方案(1)
WITH t1 AS (SELECT DISTINCT user_id, '2021-1-1' visit_date FROM UserVisits 
UNION ALL
SELECT * FROM UserVisits),

t2 AS(SELECT user_id, visit_date, ROW_NUMBER() OVER(PARTITION BY user_id ORDER BY visit_date) rn
FROM t1)

SELECT u1.user_id, MAX(DATEDIFF(u2.visit_date, u1.visit_date)) biggest_window
FROM t2 u1 JOIN t2 u2 ON u1.rn + 1 = u2.rn AND u1.user_id = u2.user_id
GROUP BY u1.user_id

-- 解决方案(2)
select user_id,max(diff) as biggest_window
from(
    select user_id,datediff(
        lead(visit_date,1,'2021-01-01') over(partition by user_id order by visit_date),visit_date
    ) as diff
    from uservisits
) tmp
group by user_id