这是一个间隙和孤岛问题。下面是使用lag()和累积sum()解决该问题的一种方法：

select min(num) num, count(*) count_num
from (
    select t.*, sum(case when num = lag_num then 0 else 1 end) over(order by id) grp
    from (
        select t.*, lag(num) over(order by id) lag_num
        from #temp t
    ) t
) t
group by grp

Demo on DB Fiddlde：

num | count_num
--: | --------:
  1 |         3
  2 |         1
  1 |         1
  2 |         2
  3 |         3

另一种方法是使用row_number

select num, count(*) 
       from (select t.*,
             (row_number() over (order by id) -
              row_number() over (partition by num order by id)
             ) as grp
             from #temp t
            ) t
group by grp, num;

DBFIDDLE

间隙和孤岛问题很有趣，因为有很多不同的方法可以解决它们。这里有一种不需要聚合的方法--尽管它需要更多地使用窗口函数。
这是可能的，因为您请求的唯一信息是计数。如果id没有间隙并且是连续的：

select num,
      lead(id, 1, max_id + 1) over (order by id) - id
from (select t.*,
             lag(num) over (order by id) as prev_num,
             max(id) over () as max_id
      from temp t
     ) t
where prev_num is null or prev_num <> num
order by id;

否则，您可以轻松地生成这样的序列：

select num,
      lead(seqnum, 1, cnt + 1) over (order by id) - seqnum
from (select t.*,
             lag(num) over (order by id) as prev_num,
             row_number() over (order by id) as seqnum,
             count(*) over () as cnt
      from temp t
     ) t
where prev_num is null or prev_num <> num
order by id;

Here是db〈〉小提琴。