这是改编自这个答案，其中包含了代码的解释。所有的改变就是增加 PARTITION BY order_id 计算每个 order_id 然后返回范围（而不是根据链接的答案返回值的总和）：

SELECT order_id,
       start_date_time,
       end_date_time
FROM   (
  SELECT order_id,
         LAG( dt ) OVER ( PARTITION BY order_id ORDER BY dt ) AS start_date_time,
         dt AS end_date_time,
         start_end
  FROM   (
    SELECT order_id,
           dt,
           CASE SUM( value ) OVER ( PARTITION BY order_id ORDER BY dt ASC, value DESC, ROWNUM ) * value
             WHEN 1 THEN 'start'
             WHEN 0 THEN 'end'
           END AS start_end
    FROM   table_name
    UNPIVOT ( dt FOR value IN ( start_date_time AS 1, end_date_time AS -1 ) )
  )
  WHERE start_end IS NOT NULL
)
WHERE  start_end = 'end';

对于你的测试数据：

CREATE TABLE table_name (
  order_id NUMBER,
  start_date_time DATE,
  end_date_time DATE
);

INSERT INTO table_name ( order_id, start_date_time, end_date_time )
SELECT 3933, TIMESTAMP '2020-02-04 08:00:00', TIMESTAMP '2020-02-04 12:00:00' FROM DUAL UNION ALL
SELECT 3933, TIMESTAMP '2020-02-04 13:30:00', TIMESTAMP '2020-02-04 17:00:00' FROM DUAL UNION ALL
SELECT 3933, TIMESTAMP '2020-02-04 14:00:00', TIMESTAMP '2020-02-04 19:00:00' FROM DUAL UNION ALL
SELECT 3933, TIMESTAMP '2020-02-05 13:40:12', TIMESTAMP '2020-02-05 14:34:48' FROM DUAL UNION ALL
SELECT 3933, TIMESTAMP '2020-02-05 14:00:00', TIMESTAMP '2020-02-05 18:55:12' FROM DUAL UNION ALL
SELECT 3933, TIMESTAMP '2020-02-05 14:49:48', TIMESTAMP '2020-02-05 15:04:48' FROM DUAL UNION ALL
SELECT 3933, TIMESTAMP '2020-02-06 08:00:00', TIMESTAMP '2020-02-06 12:00:00' FROM DUAL UNION ALL
SELECT 3933, TIMESTAMP '2020-02-06 13:30:00', TIMESTAMP '2020-02-06 17:00:00' FROM DUAL UNION ALL
SELECT 3933, TIMESTAMP '2020-02-06 14:10:12', TIMESTAMP '2020-02-06 18:49:48' FROM DUAL UNION ALL
SELECT 3933, TIMESTAMP '2020-02-07 08:00:00', TIMESTAMP '2020-02-07 10:30:00' FROM DUAL UNION ALL
SELECT 3933, TIMESTAMP '2020-02-07 08:00:00', TIMESTAMP '2020-02-07 12:00:00' FROM DUAL UNION ALL
SELECT 3933, TIMESTAMP '2020-02-07 13:30:00', TIMESTAMP '2020-02-07 17:00:00' FROM DUAL UNION ALL
SELECT 11919, TIMESTAMP '2020-05-14 09:00:00', TIMESTAMP '2020-05-14 17:00:00' FROM DUAL UNION ALL
SELECT 11919, TIMESTAMP '2020-05-14 09:00:00', TIMESTAMP '2020-05-14 17:00:00' FROM DUAL UNION ALL
SELECT 11919, TIMESTAMP '2020-05-14 15:00:00', TIMESTAMP '2020-05-14 16:30:00' FROM DUAL UNION ALL
SELECT 11919, TIMESTAMP '2020-05-15 08:40:12', TIMESTAMP '2020-05-15 16:30:00' FROM DUAL UNION ALL
SELECT 11919, TIMESTAMP '2020-05-15 09:40:12', TIMESTAMP '2020-05-15 16:30:00' FROM DUAL UNION ALL
SELECT 11919, TIMESTAMP '2020-05-15 10:15:00', TIMESTAMP '2020-05-15 12:15:00' FROM DUAL UNION ALL
SELECT 11919, TIMESTAMP '2020-05-15 13:19:48', TIMESTAMP '2020-05-15 16:00:00' FROM DUAL UNION ALL
SELECT 11919, TIMESTAMP '2020-05-18 08:49:48', TIMESTAMP '2020-05-18 09:45:00' FROM DUAL UNION ALL
SELECT 11919, TIMESTAMP '2020-05-18 10:00:00', TIMESTAMP '2020-05-18 17:00:00' FROM DUAL UNION ALL
SELECT 11919, TIMESTAMP '2020-05-18 10:00:00', TIMESTAMP '2020-05-18 16:58:12' FROM DUAL UNION ALL
SELECT 11919, TIMESTAMP '2020-05-18 15:34:48', TIMESTAMP '2020-05-18 16:10:12' FROM DUAL UNION ALL
SELECT 11919, TIMESTAMP '2020-05-18 16:30:00', TIMESTAMP '2020-05-18 16:45:00' FROM DUAL;

输出：

ORDER_ID | START_DATE_TIME     | END_DATE_TIME      
-------: | :------------------ | :------------------
    3933 | 2020-02-04 08:00:00 | 2020-02-04 12:00:00
    3933 | 2020-02-04 13:30:00 | 2020-02-04 19:00:00
    3933 | 2020-02-05 13:40:12 | 2020-02-05 18:55:12
    3933 | 2020-02-06 08:00:00 | 2020-02-06 12:00:00
    3933 | 2020-02-06 13:30:00 | 2020-02-06 18:49:48
    3933 | 2020-02-07 08:00:00 | 2020-02-07 12:00:00
    3933 | 2020-02-07 13:30:00 | 2020-02-07 17:00:00
   11919 | 2020-05-14 09:00:00 | 2020-05-14 17:00:00
   11919 | 2020-05-15 08:40:12 | 2020-05-15 16:30:00
   11919 | 2020-05-18 08:49:48 | 2020-05-18 09:45:00
   11919 | 2020-05-18 10:00:00 | 2020-05-18 17:00:00

db<>在这里摆弄

下面的解决方案使用一种称为“组的开始”方法的通用方法。
其思想是按开始日期对间隔进行排序（分别针对每个id），并按如下方式将间隔分配给组。对于每个间隔，检查其开始时间是否严格大于前面所有间隔的最大结束时间。如果是，那就开始一个新的小组。剩下的很简单-只需从每个组中选择最小开始日期和最大结束日期。
下面是如何使用分析函数实现的：

with
  has_sog_flags (order_id, start_date_time, end_date_time, flag) as (
    select order_id, start_date_time, end_date_time,
           case when start_date_time > 
                      max(end_date_time) over (partition by order_id
                                    order by start_date_time
                    rows between unbounded preceding and 1 preceding) 
                then 1 end
    from   table_name
  )
, has_groups (order_id, start_date_time, end_date_time, grp) as (
    select order_id, start_date_time, end_date_time,
           sum(flag) over (partition by order_id order by start_date_time)
    from   has_sog_flags
  )
select order_id, min(start_date_time) as start_date_time, 
       max(end_date_time) as end_date_time
from   has_groups
group  by order_id, grp
order  by order_id, start_date_time
;

一个有趣的问题是如何处理开放区间（例如 null 对于结束日期和时间，意思是“开放的未来”。查询可以相对容易地进行调整，以涵盖问题语句的此类扩展。