README.md

comments	difficulty	edit_url	tags
true	简单	https://github.com/doocs/leetcode/edit/main/solution/1000-1099/1076.Project%20Employees%20II/README.md	数据库

1076. 项目员工II 🔒

English Version

题目描述

表：Project

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| project_id  | int     |
| employee_id | int     |
+-------------+---------+
(project_id, employee_id) 是该表的主键(具有唯一值的列的组合)。
employee_id 是该表的外键(reference 列)。
该表的每一行都表明 employee_id 的雇员正在处理 Project 表中 project_id 的项目。

表：Employee

+------------------+---------+
| Column Name      | Type    |
+------------------+---------+
| employee_id      | int     |
| name             | varchar |
| experience_years | int     |
+------------------+---------+
employee_id 是该表的主键(具有唯一值的列)。
该表的每一行都包含一名雇员的信息。

 

编写一个解决方案来报告所有拥有最多员工的 项目。

以 任意顺序 返回结果表。

返回结果格式如下所示。

 

示例 1：

输入：
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                |
+-------------+--------+------------------+
输出：
+-------------+
| project_id  |
+-------------+
| 1           |
+-------------+
解释：
第一个项目有3名员工，第二个项目有2名员工。

解法

方法一

MySQL

# Write your MySQL query statement below
SELECT project_id
FROM Project
GROUP BY 1
HAVING
    COUNT(1) >= all(
        SELECT COUNT(1)
        FROM Project
        GROUP BY project_id
    );

方法二

MySQL

# Write your MySQL query statement below
WITH
    T AS (
        SELECT
            project_id,
            RANK() OVER (ORDER BY COUNT(employee_id) DESC) AS rk
        FROM Project
        GROUP BY 1
    )
SELECT project_id
FROM T
WHERE rk = 1;