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(LeetCode 185 ) 部门工资前三高的所有员工

(LeetCode 185 ) 部门工资前三高的所有员工

作者: lconcise | 来源:发表于2021-06-01 08:29 被阅读0次

    Employee 表包含所有员工信息,每个员工有其对应的工号 Id,姓名 Name,工资 Salary 和部门编号 DepartmentId 。

    +----+-------+--------+--------------+
    | Id | Name | Salary | DepartmentId |
    +----+-------+--------+--------------+
    | 1 | Joe | 85000 | 1 |
    | 2 | Henry | 80000 | 2 |
    | 3 | Sam | 60000 | 2 |
    | 4 | Max | 90000 | 1 |
    | 5 | Janet | 69000 | 1 |
    | 6 | Randy | 85000 | 1 |
    | 7 | Will | 70000 | 1 |
    +----+-------+--------+--------------+
    Department 表包含公司所有部门的信息。

    +----+----------+
    | Id | Name |
    +----+----------+
    | 1 | IT |
    | 2 | Sales |
    +----+----------+
    编写一个 SQL 查询,找出每个部门获得前三高工资的所有员工。例如,根据上述给定的表,查询结果应返回:

    +------------+----------+--------+
    | Department | Employee | Salary |
    +------------+----------+--------+
    | IT | Max | 90000 |
    | IT | Randy | 85000 |
    | IT | Joe | 85000 |
    | IT | Will | 70000 |
    | Sales | Henry | 80000 |
    | Sales | Sam | 60000 |
    +------------+----------+--------+
    解释:

    IT 部门中,Max 获得了最高的工资,Randy 和 Joe 都拿到了第二高的工资,Will 的工资排第三。销售部门(Sales)只有两名员工,Henry 的工资最高,Sam 的工资排第二。

    分析题目

    要求:
    1. 不同部门
    2. 工资前三高

    解题思路

    需要根据部门进行分组,分组后对数据进行排序。
    但是MySql 8.0 之前无法对数据进行分组,对分组后的数据进行排序
    换一种思路,同一部门,比薪资高的薪资不超过3个,那么薪资就属于前三高
    那么如果比较呢,where中关联表进行对比

    解题步骤

    1. where 中 比较,count()统计
    SELECT
        c.`name` AS department,
        a.`name` AS employee,
        a.salary 
    FROM
        employee a
        LEFT JOIN department c ON a.DepartmentId = c.id 
    WHERE
        3 > (SELECT count(b.salary) FROM employee b WHERE a.DepartmentId = b.DepartmentId AND b.salary > a.salary)
    
    image.png
    1. distinct
      统计结果并不对,并不是工资前三高,而是工资高的前三个人
    SELECT
        c.`name` AS department,
        a.`name` AS employee,
        a.salary 
    FROM
        employee a
        LEFT JOIN department c ON a.DepartmentId = c.id 
    WHERE
        3 > (SELECT count(b.salary) FROM employee b WHERE a.DepartmentId = b.DepartmentId AND b.salary > a.salary)
    

    解法二:使用MySQL8的开窗函数 dense_rank()

    select 
    Department,Employee,Salary
    from
    (
    select 
        d.name Department,e.name Employee,e.salary Salary,dense_rank()
        over(partition by d.name order by e.salary desc) rk
    from
        Employee e left join Department d
        on e.departmentid = d.id
    )t1
    where rk <= 3
    

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