[数据结构与算法]【LeetCode 数据库专项】超过经理收入的员工（181）

文章目录

• • 1. 题目
  • 2. 解法一
  • • 2.1 建表
    • 2.2 解答

1. 题目

Employee 表包含所有员工，他们的经理也属于员工。每个员工都有一个 Id ，此外还有一列对应员工的经理的 Id 。

+----+-------+--------+-----------+
| Id | Name  | Salary | ManagerId |
+----+-------+--------+-----------+
| 1  | Joe   | 70000  | 3         |
| 2  | Henry | 80000  | 4         |
| 3  | Sam   | 60000  | NULL      |
| 4  | Max   | 90000  | NULL      |
+----+-------+--------+-----------+

给定 Employee 表，编写一个 SQL 查询，该查询可以获取收入超过他们经理的员工的姓名。在上面的表格中，Joe 是唯一一个收入超过他的经理的员工。

+----------+
| Employee |
+----------+
| Joe      |
+----------+

• 来源： 力扣（LeetCode）
• 链接： https://leetcode-cn.com/problems/employees-earning-more-than-their-managers

2. 解法一

2.1 建表

Create table If Not Exists Employee (id int, name varchar(255), salary int, managerId int);
Truncate table Employee;
insert into Employee (id, name, salary, managerId) values ('1', 'Joe', '70000', '3');
insert into Employee (id, name, salary, managerId) values ('2', 'Henry', '80000', '4');
insert into Employee (id, name, salary, managerId) values ('3', 'Sam', '60000', null);
insert into Employee (id, name, salary, managerId) values ('4', 'Max', '90000', null);

2.2 解答

考虑使用联结查询，下面是不断通过添加限制条件得到具体每一步的结果。

SELECT
    *
FROM Employee AS a JOIN Employee AS b
;

+------+-------+--------+-----------+------+-------+--------+-----------+
| id   | name  | salary | managerId | id   | name  | salary | managerId |
+------+-------+--------+-----------+------+-------+--------+-----------+
|    4 | Max   |  90000 |      NULL |    1 | Joe   |  70000 |         3 |
|    3 | Sam   |  60000 |      NULL |    1 | Joe   |  70000 |         3 |
|    2 | Henry |  80000 |         4 |    1 | Joe   |  70000 |         3 |
|    1 | Joe   |  70000 |         3 |    1 | Joe   |  70000 |         3 |
|    4 | Max   |  90000 |      NULL |    2 | Henry |  80000 |         4 |
|    3 | Sam   |  60000 |      NULL |    2 | Henry |  80000 |         4 |
|    2 | Henry |  80000 |         4 |    2 | Henry |  80000 |         4 |
|    1 | Joe   |  70000 |         3 |    2 | Henry |  80000 |         4 |
|    4 | Max   |  90000 |      NULL |    3 | Sam   |  60000 |      NULL |
|    3 | Sam   |  60000 |      NULL |    3 | Sam   |  60000 |      NULL |
|    2 | Henry |  80000 |         4 |    3 | Sam   |  60000 |      NULL |
|    1 | Joe   |  70000 |         3 |    3 | Sam   |  60000 |      NULL |
|    4 | Max   |  90000 |      NULL |    4 | Max   |  90000 |      NULL |
|    3 | Sam   |  60000 |      NULL |    4 | Max   |  90000 |      NULL |
|    2 | Henry |  80000 |         4 |    4 | Max   |  90000 |      NULL |
|    1 | Joe   |  70000 |         3 |    4 | Max   |  90000 |      NULL |
+------+-------+--------+-----------+------+-------+--------+-----------+
16 rows in set (0.00 sec)

SELECT
    *
FROM Employee AS a JOIN Employee AS b
    ON a.ManagerId = b.Id
    AND a.Salary > b.Salary
;

+------+------+--------+-----------+------+------+--------+-----------+
| id   | name | salary | managerId | id   | name | salary | managerId |
+------+------+--------+-----------+------+------+--------+-----------+
|    1 | Joe  |  70000 |         3 |    3 | Sam  |  60000 |      NULL |
+------+------+--------+-----------+------+------+--------+-----------+
1 row in set (0.00 sec)

SELECT
    a.Name AS 'Employee'
FROM Employee AS a JOIN Employee AS b
    ON a.ManagerId = b.Id
    AND a.Salary > b.Salary
;

+----------+
| Employee |
+----------+
| Joe      |
+----------+
1 row in set (0.00 sec)