README.md

comments	difficulty	edit_url	tags
true	简单	https://github.com/doocs/leetcode/edit/main/solution/0500-0599/0586.Customer%20Placing%20the%20Largest%20Number%20of%20Orders/README.md	数据库

586. 订单最多的客户

English Version

题目描述

表: Orders

+-----------------+----------+
| Column Name     | Type     |
+-----------------+----------+
| order_number    | int      |
| customer_number | int      |
+-----------------+----------+
在 SQL 中，Order_number是该表的主键。
此表包含关于订单ID和客户ID的信息。

 

查找下了 最多订单 的客户的 customer_number 。

测试用例生成后， 恰好有一个客户 比任何其他客户下了更多的订单。

查询结果格式如下所示。

 

示例 1:

输入: 
Orders 表:
+--------------+-----------------+
| order_number | customer_number |
+--------------+-----------------+
| 1            | 1               |
| 2            | 2               |
| 3            | 3               |
| 4            | 3               |
+--------------+-----------------+
输出: 
+-----------------+
| customer_number |
+-----------------+
| 3               |
+-----------------+
解释: 
customer_number 为 '3' 的顾客有两个订单，比顾客 '1' 或者 '2' 都要多，因为他们只有一个订单。
所以结果是该顾客的 customer_number ，也就是 3 。

 

进阶： 如果有多位顾客订单数并列最多，你能找到他们所有的 customer_number 吗？

解法

方法一：分组 + 排序

我们可以使用 GROUP BY 将数据按照 customer_number 进行分组，然后按照 count(1) 进行降序排序，最后取第一条记录的 customer_number 即可。

MySQL

# Write your MySQL query statement below
SELECT
    customer_number
FROM orders
GROUP BY customer_number
ORDER BY COUNT(1) DESC
LIMIT 1;

方法二

MySQL

/* Write your T-SQL query statement below */
SELECT TOP 1
    customer_number
FROM
    orders
GROUP BY customer_number
ORDER BY COUNT(customer_number) DESC;