目录
- [1 题目](#1 题目)
- [2 建表语句](#2 建表语句)
- [3 题解](#3 题解)
题目来源：拼多多。
1 题目
有一张表t_id记录了id，id不重复，但是会存在间断，求出连续段的起始位置和结束位置。
样例数据
+-----+
| id |
+-----+
| 1 |
| 2 |
| 3 |
| 5 |
| 6 |
| 8 |
| 10 |
| 12 |
| 13 |
| 14 |
| 15 |
+-----+2 建表语句
sql
--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)3 题解
（1）lag()函数进行开窗计算与上一行的差值；
sql
select id,
id - lag(id) over (order by id) as diff
from t_id执行结果
+-----+-------+
| id | diff |
+-----+-------+
| 1 | NULL |
| 2 | 1 |
| 3 | 1 |
| 5 | 2 |
| 6 | 1 |
| 8 | 2 |
| 10 | 2 |
| 12 | 2 |
| 13 | 1 |
| 14 | 1 |
| 15 | 1 |
+-----+-------+（2）获得分组字段
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。
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执行结果
+-----+-------------+
| 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 |
+-----+-------------+（3）得出结果
sql
select group_type,
min(id) as start_pos,
max(id) as end_pos
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执行结果
+-------------+------------+----------+
| group_type | start_pos | end_pos |
+-------------+------------+----------+
| 1 | 1 | 3 |
| 2 | 5 | 6 |
| 3 | 8 | 8 |
| 4 | 10 | 10 |
| 5 | 12 | 15 |
+-------------+------------+----------+