这是一种缺口和孤岛问题。您可以使用累计值来确定每一行是否是连续时间段的开始 max() . 然后用一个累积的和来得到时间段的开始。最后是聚合。因此，对于各个时段：

select name, min(start) as start, max(finish) as finish
from (select t.*,
             sum(case when prev_finish >= start then 0 else 1 end) over (partition by name order by start) as grp
      from (select t.*,
                   max(finish) over (partition by name order by start rows between unbounded preceding and 1 preceding) as prev_finish
            from t
           ) t
     ) t
group by name, grp;

然后可以执行最终聚合：

with cte as (
      Select 'Marc' as name, '2020-01-01 10:00:00.000' as start,'2020-01-01 11:00:00.000' as finish
      union all
      Select 'Marc' as name, '2020-01-01 10:15:00.000' as start,'2020-01-01 11:00:00.000' as finish
      union all
      Select 'Marc' as name, '2020-01-01 12:10:00.000' as start,'2020-01-01 14:00:00.000' as finish
     ),
     t as (
      select name, min(start) as start, max(finish) as finish
      from (select t.*,
                   sum(case when prev_finish >= start then 0 else 1 end) over (partition by name order by start) as grp
            from (select t.*,
                         max(finish) over (partition by name order by start rows between unbounded preceding and 1 preceding) as prev_finish
                  from cte t
                 ) t
           ) t
      group by name, grp
     ) 
select name, sum(datediff(hour, start, finish)) as hours
from t
group by name;

这是一把小提琴。