一、题目
有一张表t_id记录了id,id不重复,但是会存在间断,求出连续段的最后一个数及每个连续段的个数。
+-----+
| id |
+-----+
| 1 |
| 2 |
| 3 |
| 5 |
| 6 |
| 8 |
| 10 |
| 12 |
| 13 |
| 14 |
| 15 |
+-----+
维度 | 评分 |
---|---|
题目难度 | ⭐️⭐️⭐️⭐️ |
题目清晰度 | ⭐️⭐️⭐️⭐️⭐️ |
业务常见度 | ⭐️⭐️⭐️ |
执行SQL
select id,
id - lag(id) over (order by id) as diff
from t_id
查询结果
+-----+-------+
| id | diff |
+-----+-------+
| 1 | NULL |
| 2 | 1 |
| 3 | 1 |
| 5 | 2 |
| 6 | 1 |
| 8 | 2 |
| 10 | 2 |
| 12 | 2 |
| 13 | 1 |
| 14 | 1 |
| 15 | 1 |
+-----+-------+
根据diff进行判断,如果差值为1代表连续赋值为0,否则代表不连续赋值为1,然后使用sum()进行累积计算,获得分组依据字段。
执行SQL
select id,
sum(if(diff = 1, 0, 1)) over (order by id) as group_type
from (select id,
id - lag(id) over (order by id) as diff
from t_id) t
查询结果
+-----+-------------+
| id | group_type |
+-----+-------------+
| 1 | 1 |
| 2 | 1 |
| 3 | 1 |
| 5 | 2 |
| 6 | 2 |
| 8 | 3 |
| 10 | 4 |
| 12 | 5 |
| 13 | 5 |
| 14 | 5 |
| 15 | 5 |
+-----+-------------+
执行SQL
select group_type,
max(id) as max_part,
count(1) as num_part
from (select id,
sum(if(diff = 1, 0, 1)) over (order by id) as group_type
from (select id,
id - lag(id) over (order by id) as diff
from t_id) t) tt
group by group_type
查询结果
+-------------+-----------+-----------+
| group_type | max_part | num_part |
+-------------+-----------+-----------+
| 1 | 3 | 3 |
| 2 | 6 | 2 |
| 3 | 8 | 1 |
| 4 | 10 | 1 |
| 5 | 15 | 4 |
+-------------+-----------+-----------+
--建表语句
CREATE TABLE t_id (
id bigint COMMENT 'ID'
) COMMENT 'ID记录表'
ROW FORMAT DELIMITED FIELDS TERMINATED BY '\t'
;
-- 插入数据
insert into t_id(id)
values
(1),
(2),
(3),
(5),
(6),
(8),
(10),
(12),
(13),
(14),
(15)