试试这个。

SELECT id, ( max( TO_DATE ( time,'YYYYMMDDHHMI' ) )
- min( TO_DATE ( time,'YYYYMMDDHHMI' ) ) ) *24*60 diff_in_minutes from 
(
select t.*,
row_number()   OVER ( ORDER BY 
                    TO_DATE ( time,'YYYYMMDDHHMI' ) )
- row_number() OVER ( PARTITION BY ID ORDER BY 
                    TO_DATE ( time,'YYYYMMDDHHMI' ) ) seq
FROM Table1 t ORDER BY time
  ) GROUP BY ID,seq
  ORDER BY max(time)
  ;

演示
编辑：这个答案是考虑到op已经标记了 oracle 。现在改为 hive .
作为 hive 的替代品 TO_DATE 在甲骨文中，

unix_timestamp(time, 'yyyyMMddhhmm')

可以使用。

我建议一些转变：
添加一个指示行是否是其组中的第一行（标记为1，否则为空）
计算行前面的此类标志的数目以了解其组号
然后你就可以按新的组号分组了。

oracle版本（原始问题）

with q1 as (
    select to_date(time, 'YYYYMMDDHH24MI') time, id, 
           case id when lag(id) over(order by time) then null else 1 end first_in_group 
    from t
), q2 as (
    select time, id, count(first_in_group) over (order by time) grp_id
    from   q1
)
select   min(id) id, (max(time) - min(time)) * 24 * 60 minutes
from     q2
group by grp_id
order by grp_id

sql小提琴

配置单元版本

不同的数据库引擎使用不同的函数来处理日期/时间值，所以使用hive的 unix_timestamp 处理它返回的秒数：

with q1 as (
    select unix_timestamp(time, 'yyyyMMddHHmm')/60 time, id, 
           case id when lag(id) over(order by time) then null else 1 end first_in_group 
    from t
), q2 as (
    select time, id, count(first_in_group) over (order by time) grp_id
    from   q1
)
select   min(id) id, max(time) - min(time) minutes
from     q2
group by grp_id
order by grp_id