有意思的SQL(2)

求教第二行的入院时间减去第一行的出院时间SQL怎样写？

如图都是同一个人,我想计算一个时间差，第二行的入院时间减去第一行的出院时间，第三行的入院时间减去第二行的出院时间，以此类推。
请教这种SQL能实现吗？
原问题
https://www.modb.pro/issue/13644

解法一

with cte1 as
(
select 1 as id, '张三'  as sname, '2022-01-01' as bdate,'2022-01-03' as edate 
union all
select 2 as id,'张三'  as sname, '2022-01-08' as bdate,'2022-01-10' as edate 
union all
select 3 as id,'张三'  as sname, '2022-01-17' as bdate,'2022-01-21' as edate 
union all
select 4 as id,'张三'  as sname, '2022-01-24' as bdate,'2022-01-25' as edate 
union all
select 5 as id,'张三'  as sname, '2022-01-27' as bdate,'2022-01-29' as edate 
union all
select 6 as id,'张三'  as sname, '2022-02-03' as bdate,'2022-02-10' as edate 
union all
select 7 as id, '张三'  as sname, '2022-02-15' as bdate,'2022-02-23' as edate 
)
select id,sname,bdate,edate, nextbdate,datediff(nextbdate,edate) as days from 
(
select *,last_value(bdate) over(order by id ROWS BETWEEN CURRENT ROW AND 1 FOLLOWING)  as nextbdate
 from cte1
) as t1

解法二

with cte1 as
(
select 1 as id, '张三'  as sname, '2022-01-01' as bdate,'2022-01-03' as edate 
union all
select 2 as id,'张三'  as sname, '2022-01-08' as bdate,'2022-01-10' as edate 
union all
select 3 as id,'张三'  as sname, '2022-01-17' as bdate,'2022-01-21' as edate 
union all
select 4 as id,'张三'  as sname, '2022-01-24' as bdate,'2022-01-25' as edate 
union all
select 5 as id,'张三'  as sname, '2022-01-27' as bdate,'2022-01-29' as edate 
union all
select 6 as id,'张三'  as sname, '2022-02-03' as bdate,'2022-02-10' as edate 
union all
select 7 as id, '张三'  as sname, '2022-02-15' as bdate,'2022-02-23' as edate 
)
select id,sname,bdate,edate, datediff(nextbdate,edate) as days from 
(
select *,lead(bdate) over(order by id)  as nextbdate
 from cte1
) as t1