有奖问答|LeetCode 181. 超过经理收入的员工--讨论贴

2023-09-11 06:03:01

题目

《超过经理收入的员工》算法题如下：

给定一个包含员工姓名、经理姓名和薪资的表，查找所有收入高于其经理的员工。

表名为Employee，包含以下列：

id：员工的ID，唯一标识符
name：员工的姓名
managerId：经理的ID，如果员工是经理，则此列为NULL
salary：员工的薪资

例如，给定以下Employee表：

id	name	managerId	salary
1	John Doe	NULL	10000
2	Jane Smith	1	8000
3	Michael Jones	2	7000
4	Mary Johnson	1	6000
5	Robert Brown	3	5000

查询结果应为：

name	salary
John Doe	10000

解题思路

这道算法题可以使用SQL语句来解决。我们可以使用以下步骤来实现：

首先，我们需要找到所有经理的ID。我们可以使用以下SQL语句：

SELECT DISTINCT managerId FROM Employee WHERE managerId IS NOT NULL;

然后，我们可以使用这些经理的ID来查找所有收入高于其经理的员工。我们可以使用以下SQL语句：

SELECT E1.name, E1.salary
FROM Employee AS E1
JOIN Employee AS E2 ON E1.managerId = E2.id
WHERE E1.salary > E2.salary;

实现方法

我们可以使用以下Python代码来实现上述算法：

import mysql.connector

def find_employees_with_higher_salary_than_manager(db_host, db_user, db_password, db_name):
  """
  Finds all employees with a higher salary than their manager.

  Args:
    db_host: The hostname of the MySQL database.
    db_user: The username to connect to the MySQL database.
    db_password: The password to connect to the MySQL database.
    db_name: The name of the MySQL database.

  Returns:
    A list of tuples containing the names and salaries of employees with a higher salary than their manager.
  """

  # Connect to the MySQL database.
  connection = mysql.connector.connect(
      host=db_host,
      user=db_user,
      password=db_password,
      database=db_name
  )

  # Create a cursor object.
  cursor = connection.cursor()

  # Find all managers' IDs.
  cursor.execute("SELECT DISTINCT managerId FROM Employee WHERE managerId IS NOT NULL;")
  manager_ids = [row[0] for row in cursor.fetchall()]

  # Find all employees with a higher salary than their manager.
  cursor.execute("""
    SELECT E1.name, E1.salary
    FROM Employee AS E1
    JOIN Employee AS E2 ON E1.managerId = E2.id
    WHERE E1.salary > E2.salary;
  """)
  employees = cursor.fetchall()

  # Close the cursor and connection.
  cursor.close()
  connection.close()

  return employees


if __name__ == "__main__":
  # Example usage.
  employees = find_employees_with_higher_salary_than_manager(
      db_host="localhost",
      db_user="root",
      db_password="password",
      db_name="company"
  )
  for employee in employees:
    print(f"{employee[0]} earns ${employee[1]}")