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feat: add sql solutions to lc problems: No.2853,2854 (doocs#1634)
* No.2853.Highest Salaries Difference * No.2854.Rolling Average Steps
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# [2853. Highest Salaries Difference](https://leetcode.cn/problems/highest-salaries-difference)
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[English Version](/solution/2800-2899/2853.Highest%20Salaries%20Difference/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>Table: <code><font face="monospace">Salaries</font></code></p>
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<pre>
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+-------------+---------+
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| Column Name | Type |
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+-------------+---------+
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| emp_name | varchar |
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| department | varchar |
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| salary | int |
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+-------------+---------+
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(emp_name, department) is the primary key for this table.
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Each row of this table contains emp_name, department and salary. There will be <strong>at least one</strong> entry for the engineering and marketing departments.
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</pre>
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<p>Write an SQL query to calculate the difference between the <strong>highest</strong> salaries in the <strong>marketing</strong> and <strong>engineering</strong> <code>department</code>. Output the absolute difference in salaries.</p>
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<p>Return<em> the result table.</em></p>
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<p>The query result format is in the following example.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong>
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Salaries table:
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+----------+-------------+--------+
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| emp_name | department | salary |
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+----------+-------------+--------+
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| Kathy | Engineering | 50000 |
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| Roy | Marketing | 30000 |
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| Charles | Engineering | 45000 |
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| Jack | Engineering | 85000 |
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| Benjamin | Marketing | 34000 |
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| Anthony | Marketing | 42000 |
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| Edward | Engineering | 102000 |
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| Terry | Engineering | 44000 |
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| Evelyn | Marketing | 53000 |
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| Arthur | Engineering | 32000 |
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+----------+-------------+--------+
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<strong>Output:</strong>
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+-------------------+
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| salary_difference |
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+-------------------+
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| 49000 |
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+-------------------+
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<strong>Explanation:</strong>
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- The Engineering and Marketing departments have the highest salaries of 102,000 and 53,000, respectively. Resulting in an absolute difference of 49,000.
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</pre>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:GROUP BY 分组**
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我们可以先分别计算出每个部门的最高工资,然后再计算两个最高工资的差值。
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<!-- tabs:start -->
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### **SQL**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```sql
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# Write your MySQL query statement below
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SELECT max(s) - min(s) AS salary_difference
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FROM
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(
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SELECT max(salary) AS s
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FROM Salaries
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GROUP BY department
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) AS t;
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```
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<!-- tabs:end -->
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# [2853. Highest Salaries Difference](https://leetcode.com/problems/highest-salaries-difference)
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[中文文档](/solution/2800-2899/2853.Highest%20Salaries%20Difference/README.md)
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## Description
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<p>Table: <code><font face="monospace">Salaries</font></code></p>
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<pre>
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+-------------+---------+
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| Column Name | Type |
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+-------------+---------+
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| emp_name | varchar |
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| department | varchar |
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| salary | int |
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+-------------+---------+
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(emp_name, department) is the primary key for this table.
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Each row of this table contains emp_name, department and salary. There will be <strong>at least one</strong> entry for the engineering and marketing departments.
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</pre>
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<p>Write an SQL query to calculate the difference between the <strong>highest</strong> salaries in the <strong>marketing</strong> and <strong>engineering</strong> <code>department</code>. Output the absolute difference in salaries.</p>
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<p>Return<em> the result table.</em></p>
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<p>The query result format is in the following example.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong>
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Salaries table:
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+----------+-------------+--------+
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| emp_name | department | salary |
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+----------+-------------+--------+
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| Kathy | Engineering | 50000 |
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| Roy | Marketing | 30000 |
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| Charles | Engineering | 45000 |
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| Jack | Engineering | 85000 |
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| Benjamin | Marketing | 34000 |
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| Anthony | Marketing | 42000 |
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| Edward | Engineering | 102000 |
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| Terry | Engineering | 44000 |
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| Evelyn | Marketing | 53000 |
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| Arthur | Engineering | 32000 |
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+----------+-------------+--------+
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<strong>Output:</strong>
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+-------------------+
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| salary_difference |
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+-------------------+
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| 49000 |
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+-------------------+
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<strong>Explanation:</strong>
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- The Engineering and Marketing departments have the highest salaries of 102,000 and 53,000, respectively. Resulting in an absolute difference of 49,000.
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</pre>
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## Solutions
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<!-- tabs:start -->
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### **SQL**
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```sql
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# Write your MySQL query statement below
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SELECT max(s) - min(s) AS salary_difference
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FROM
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(
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SELECT max(salary) AS s
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FROM Salaries
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GROUP BY department
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) AS t;
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```
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<!-- tabs:end -->
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# Write your MySQL query statement below
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SELECT max(s) - min(s) AS salary_difference
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FROM
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(
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SELECT max(salary) AS s
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FROM Salaries
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GROUP BY department
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) AS t;
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# [2854. Rolling Average Steps](https://leetcode.cn/problems/rolling-average-steps)
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[English Version](/solution/2800-2899/2854.Rolling%20Average%20Steps/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>Table: <code><font face="monospace">Steps</font></code></p>
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<pre>
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+-------------+------+
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| Column Name | Type |
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+-------------+------+
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| user_id | int |
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| steps_count | int |
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| steps_date | date |
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+-------------+------+
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(user_id, steps_date) is the primary key for this table.
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Each row of this table contains user_id, steps_count, and steps_date.
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</pre>
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<p>Write a solution to calculate <code>3-day</code> <strong>rolling averages</strong> of steps for each user.</p>
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<p>We calculate the <code>n-day</code> <strong>rolling average</strong> this way:</p>
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<ul>
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<li>For each day, we calculate the average of <code>n</code> consecutive days of step counts ending on that day if available, otherwise, <code>n-day</code> rolling average is not defined for it.</li>
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</ul>
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<p>Output the <code>user_id</code>, <code>steps_date</code>, and rolling average. Round the rolling average to <strong>two decimal places</strong>.</p>
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<p>Return<em> the result table ordered by </em><code>user_id</code><em>, </em><code>steps_date</code><em> in <strong>ascending</strong> order.</em></p>
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<p>The result format is in the following example.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong>
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Steps table:
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+---------+-------------+------------+
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| user_id | steps_count | steps_date |
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+---------+-------------+------------+
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| 1 | 687 | 2021年09月02日 |
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| 1 | 395 | 2021年09月04日 |
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| 1 | 499 | 2021年09月05日 |
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| 1 | 712 | 2021年09月06日 |
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| 1 | 576 | 2021年09月07日 |
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| 2 | 153 | 2021年09月06日 |
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| 2 | 171 | 2021年09月07日 |
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| 2 | 530 | 2021年09月08日 |
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| 3 | 945 | 2021年09月04日 |
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| 3 | 120 | 2021年09月07日 |
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| 3 | 557 | 2021年09月08日 |
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| 3 | 840 | 2021年09月09日 |
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| 3 | 627 | 2021年09月10日 |
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| 5 | 382 | 2021年09月05日 |
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| 6 | 480 | 2021年09月01日 |
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| 6 | 191 | 2021年09月02日 |
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| 6 | 303 | 2021年09月05日 |
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+---------+-------------+------------+
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<strong>Output:</strong>
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+---------+------------+-----------------+
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| user_id | steps_date | rolling_average |
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+---------+------------+-----------------+
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| 1 | 2021年09月06日 | 535.33 |
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| 1 | 2021年09月07日 | 595.67 |
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| 2 | 2021年09月08日 | 284.67 |
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| 3 | 2021年09月09日 | 505.67 |
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| 3 | 2021年09月10日 | 674.67 |
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+---------+------------+-----------------+
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<strong>Explanation:</strong>
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- For user id 1, the step counts for the three consecutive days up to 2021年09月06日 are available. Consequently, the rolling average for this particular date is computed as (395 + 499 + 712) / 3 = 535.33.
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- For user id 1, the step counts for the three consecutive days up to 2021年09月07日 are available. Consequently, the rolling average for this particular date is computed as (499 + 712 + 576) / 3 = 595.67.
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- For user id 2, the step counts for the three consecutive days up to 2021年09月08日 are available. Consequently, the rolling average for this particular date is computed as (153 + 171 + 530) / 3 = 284.67.
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- For user id 3, the step counts for the three consecutive days up to 2021年09月09日 are available. Consequently, the rolling average for this particular date is computed as (120 + 557 + 840) / 3 = 505.67.
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- For user id 3, the step counts for the three consecutive days up to 2021年09月10日 are available. Consequently, the rolling average for this particular date is computed as (557 + 840 + 627) / 3 = 674.67.
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- For user id 4 and 5, the calculation of the rolling average is not viable as there is insufficient data for the consecutive three days. Output table ordered by user_id and steps_date in ascending order.</pre>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:窗口函数**
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我们用窗口函数 `LAG() OVER()` 来计算每个用户当前日期与上上个日期之间的天数差,如果为 2ドル,ドル说明这两个日期之间有连续 3ドル$ 天的数据,我们可以利用窗口函数 `AVG() OVER()` 来计算这 3ドル$ 个数据的平均值。
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<!-- tabs:start -->
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### **SQL**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```sql
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# Write your MySQL query statement below
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WITH
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T AS (
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SELECT
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user_id,
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steps_date,
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round(
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avg(steps_count) OVER (
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PARTITION BY user_id
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ORDER BY steps_date
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ROWS 2 PRECEDING
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),
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2
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) AS rolling_average,
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datediff(
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steps_date,
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lag(steps_date, 2) OVER (
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PARTITION BY user_id
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ORDER BY steps_date
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)
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) = 2 AS st
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FROM Steps
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)
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SELECT
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user_id,
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steps_date,
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rolling_average
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FROM T
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WHERE st = 1
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ORDER BY 1, 2;
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```
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<!-- tabs:end -->

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