LeetCode-1076. 项目员工II（简单）

Table: Project

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| project_id  | int     |
| employee_id | int     |
+-------------+---------+
主键为 (project_id, employee_id)。
employee_id 是员工表 Employee 表的外键。
Table: Employee

+------------------+---------+
| Column Name      | Type    |
+------------------+---------+
| employee_id      | int     |
| name             | varchar |
| experience_years | int     |
+------------------+---------+
主键是 employee_id。
 

编写一个SQL查询，报告所有雇员最多的项目。

查询结果格式如下所示：

Project table:
+-------------+-------------+
| project_id  | employee_id |
+-------------+-------------+
| 1           | 1           |
| 1           | 2           |
| 1           | 3           |
| 2           | 1           |
| 2           | 4           |
+-------------+-------------+

Employee table:
+-------------+--------+------------------+
| employee_id | name   | experience_years |
+-------------+--------+------------------+
| 1           | Khaled | 3                |
| 2           | Ali    | 2                |
| 3           | John   | 1                |
| 4           | Doe    | 2                |
+-------------+--------+------------------+

Result table:
+-------------+
| project_id  |
+-------------+
| 1           |
+-------------+
第一个项目有3名员工，第二个项目有2名员工。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/project-employees-ii
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审题：查询雇员最多的项目

思考：按照项目表，项目分组，排序，查找第一个，需要判断项目人数一样多的问题。

解题：

解法一

先算出每个项目的员工数。应用group by分组。

(
    SELECT P.project_id,count(distinct P.employee_id) as `cnt`
    from Project as P
    group by P.project_id
) AS A

再求所有项目中最多的人数。

在上面基础上，降序排，取第一个。

(
    SELECT count(distinct P.employee_id) as `cnt`
    from Project as P
    group by P.project_id
    order by cnt desc
    limit 0,1
) AS B

连接表A和表B，选出员工数等于最大值的项目。

SELECT A.project_id
FROM 
(
    SELECT P.project_id,count(distinct P.employee_id) as `cnt`
    from Project as P
    group by P.project_id
) AS A
JOIN 
(
    SELECT count(distinct P.employee_id) as `cnt`
    from Project as P
    group by P.project_id
    order by cnt desc
    limit 0,1
) AS B
	ON(A.cnt = B.cnt)
GROUP BY A.project_id

解法二

先求最多的人数。

逻辑同解法一。

(
    SELECT count(distinct P.employee_id) as `cnt`
    from Project as P
    group by P.project_id
    order by cnt desc
    limit 0,1
) AS B

项目表与表B叉积，对项目分组，计算每个组的人数，选出每组人数等于最多人数的行。

SELECT A.project_id
from Project AS A,
(
    SELECT count(distinct P.employee_id) as `cnt`
    from Project as P
    group by P.project_id
    order by cnt desc
    limit 0,1
) AS B
GROUP BY A.project_id
having COUNT(distinct A.employee_id) = MAX(B.cnt)

注意：having子句中用到MAX(B.cnt)。由于group by只对A.project_id分组。B.cnt不能直接用在“=”之后，尽管每行的B.cnt相等。需要将B.cnt包装在聚集函数中。此处选了MAX。或者将group by改为对A.project_id和B.cnt分组，B.cnt就可以直接用在having子句中。

SELECT A.project_id
from Project AS A,
(
    SELECT count(distinct P.employee_id) as `cnt`
    from Project as P
    group by P.project_id
    order by cnt desc
    limit 0,1
) AS B
GROUP BY A.project_id,B.cnt
having COUNT(distinct A.employee_id) = B.cnt

知识点：