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‎problem/0183.md

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---
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title: 183. 从不订购的客户
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description: LeetCode 183. 从不订购的客户题解,Customers Who Never Order,包含解题思路、复杂度分析以及完整的 JavaScript 代码实现。
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keywords:
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- LeetCode
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- 183. 从不订购的客户
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- 从不订购的客户
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- Customers Who Never Order
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- 解题思路
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- 数据库
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---
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# 183. 从不订购的客户
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🟢 <font color=#15bd66>Easy</font>&emsp; 🔖&ensp; [`数据库`](/tag/database.md)&emsp; 🔗&ensp;[`力扣`](https://leetcode.cn/problems/customers-who-never-order) [`LeetCode`](https://leetcode.com/problems/customers-who-never-order)
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## 题目
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Table: `Customers`
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> ```
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> +-------------+---------+
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> | Column Name | Type |
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> +-------------+---------+
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> | id | int |
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> | name | varchar |
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> +-------------+---------+
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> ```
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>
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> id is the primary key (column with unique values) for this table.
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>
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> Each row of this table indicates the ID and name of a customer.
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Table: `Orders`
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> ```
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> +-------------+------+
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> | Column Name | Type |
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> +-------------+------+
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> | id | int |
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> | customerId | int |
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> +-------------+------+
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> ```
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>
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> id is the primary key (column with unique values) for this table.
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>
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> customerId is a foreign key (reference columns) of the ID from the Customers table.
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>
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> Each row of this table indicates the ID of an order and the ID of the customer who ordered it.
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Write a solution to find all customers who never order anything.
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Return the result table in **any order**.
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The result format is in the following example.
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**Example 1:**
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> Input:
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>
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> Customers table:
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>
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> ```
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> +----+-------+
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> | id | name |
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> +----+-------+
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> | 1 | Joe |
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> | 2 | Henry |
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> | 3 | Sam |
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> | 4 | Max |
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> +----+-------+
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> ```
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>
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> Orders table:
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>
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> ```
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> +----+------------+
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> | id | customerId |
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> +----+------------+
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> | 1 | 3 |
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> | 2 | 1 |
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> +----+------------+
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> ```
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>
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> Output:
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>
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> ```
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> +-----------+
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> | Customers |
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> +-----------+
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> | Henry |
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> | Max |
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> +-----------+
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> ```
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## 题目大意
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`Customers` 表:
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> ```
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> +-------------+---------+
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> | Column Name | Type |
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> +-------------+---------+
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> | id | int |
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> | name | varchar |
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> +-------------+---------+
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> ```
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>
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> 在 SQL 中,id 是该表的主键。
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>
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> 该表的每一行都表示客户的 ID 和名称。
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`Orders` 表:
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> ```
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> +-------------+------+
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> | Column Name | Type |
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> +-------------+------+
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> | id | int |
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> | customerId | int |
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> +-------------+------+
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> ```
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>
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> 在 SQL 中,id 是该表的主键。
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>
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> customerId 是 Customers 表中 ID 的外键( Pandas 中的连接键)。
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>
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> 该表的每一行都表示订单的 ID 和订购该订单的客户的 ID。
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找出所有从不点任何东西的顾客。
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以 **任意顺序** 返回结果表。
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结果格式如下所示。
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**示例 1:**
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> **输入:**
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>
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> Customers table:
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>
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> ```
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> +----+-------+
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> | id | name |
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> +----+-------+
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> | 1 | Joe |
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> | 2 | Henry |
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> | 3 | Sam |
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> | 4 | Max |
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> +----+-------+
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> ```
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>
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> Orders table:
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>
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> ```
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> +----+------------+
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> | id | customerId |
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> +----+------------+
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> | 1 | 3 |
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> | 2 | 1 |
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> +----+------------+
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> ```
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>
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> **输出:**
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>
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> ```
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> +-----------+
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> | Customers |
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> +-----------+
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> | Henry |
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> | Max |
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> +-----------+
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> ```
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## 解题思路
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### MySQL
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1. **返回字段**:
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`SELECT` 指定要返回的字段:`name`(客户名称),重命名为 `Customers`。
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`FROM Customers` 表作为数据来源。
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2. **筛选条件**:
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- 使用 `LEFT JOIN` 将 `Customers` 表与 `Orders` 表连接,关联条件是 `Customers.id = Orders.customerId`。
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- 筛选出未下单的客户,即 `Orders.customerId` 为 `NULL` 的记录。
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- 使用 `WHERE o.customerId IS NULL` 筛选条件。
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#### 复杂度分析
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- **时间复杂度**:`O(n + m)`,其中 `n` 是 `Customers` 表的记录数,`m` 是 `Orders` 表的记录数。
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- **空间复杂度**:返回的结果占用的空间与未下单的客户数成正比。
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---
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### Pandas
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1. **加载数据**:
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将 `Customers` 和 `Orders` 数据加载到 Pandas 的两个 `DataFrame` 中。
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2. **左连接**:
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使用 `merge` 模拟 SQL 的 `LEFT JOIN`,`Customers` 为左表,`Orders` 为右表,关联条件是 `Customers.id = Orders.customerId`。
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3. **筛选条件**:
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筛选出 `customerId` 为 `NaN` 的记录,即未下过订单的客户。
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4. **选择列**:
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保留筛选后的 `name` 列,并重命名为 `Customers` ,作为最终结果。
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## 代码
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::: code-tabs
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@tab MySQL
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```sql
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SELECT c.name as Customers
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FROM Customers c
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LEFT JOIN Orders o
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ON c.id = o.customerId
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WHERE o.customerId IS NULL
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```
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@tab Pandas
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```python
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import pandas as pd
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def customers_without_orders(customers: pd.DataFrame, orders: pd.DataFrame) -> pd.DataFrame:
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# LEFT JOIN 模拟
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merged = pd.merge(customers, orders, left_on='id', right_on='customerId', how='left')
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# 筛选未下订单的客户
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return merged[merged['customerId'].isna()][['name']].rename(columns={'name': 'Customers'})
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```
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:::

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