leetcode刷题笔记一百八十四题 && 一百八十五题
leetcode刷题笔记一百八十四题 && 一百八十五题
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184问题描述:
Employee 表包含所有员工信息,每个员工有其对应的 Id, salary 和 department Id。
+----+-------+--------+--------------+
| Id | Name | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1 | Joe | 70000 | 1 |
| 2 | Jim | 90000 | 1 |
| 3 | Henry | 80000 | 2 |
| 4 | Sam | 60000 | 2 |
| 5 | Max | 90000 | 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,找出每个部门工资最高的员工。对于上述表,您的 SQL 查询应返回以下行(行的顺序无关紧要)。+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT | Max | 90000 |
| IT | Jim | 90000 |
| Sales | Henry | 80000 |
+------------+----------+--------+
解释:Max 和 Jim 在 IT 部门的工资都是最高的,Henry 在销售部的工资最高。
# Write your MySQL query statement below
SELECT B.name AS 'Department',
A.name AS 'Employee',
Salary FROM Employee AS A JOIN Department B ON A.DepartmentId = B.Id
WHERE (A.DepartmentId , A.Salary) IN (
SELECT DepartmentId, MAX(Salary) FROM Employee GROUP BY DepartmentId
);
185问题描述:
SQL架构
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。
# Write your MySQL query statement below
SELECT B.Name AS Department, A.Name AS Employee,A.Salary FROM
(SELECT dense_rank() OVER(PARTITION BY DepartmentId ORDER BY Salary DESC) AS RANKId, Name, Salary, DepartmentId FROM Employee) AS A JOIN Department AS B
ON A.DepartmentId = B.Id WHERE A.RANKId <= 3;