3580. 寻找持续进步的员工 （单调性的模板题）

3580. 寻找持续进步的员工 - 力扣（LeetCode）

题目描述

表：employees

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| employee_id | int     |
| name        | varchar |
+-------------+---------+
employee_id 是这张表的唯一主键。
每一行包含一名员工的信息。

表：performance_reviews

+-------------+------+
| Column Name | Type |
+-------------+------+
| review_id   | int  |
| employee_id | int  |
| review_date | date |
| rating      | int  |
+-------------+------+
review_id 是这张表的唯一主键。
每一行表示一名员工的绩效评估。评分在 1-5 的范围内，5分代表优秀，1分代表较差。

编写一个解决方案，以找到在过去三次评估中持续提高绩效的员工。

• 员工 至少需要 3 次评估 才能被考虑
• 员工过去的 3 次评估，评分必须 严格递增（每次评价都比上一次好）
• 根据 review_date 为每位员工分析最近的 3 次评估
• 进步分数 为最后 3 次评估中最后一次评分与最早一次评分之间的差值

返回结果表以 进步分数 降序 排序，然后以 名字 升序 排序。

结果格式如下所示。

示例：

输入：

employees 表：

+-------------+----------------+
| employee_id | name           |
+-------------+----------------+
| 1           | Alice Johnson  |
| 2           | Bob Smith      |
| 3           | Carol Davis    |
| 4           | David Wilson   |
| 5           | Emma Brown     |
+-------------+----------------+

performance_reviews 表：

+-----------+-------------+-------------+--------+
| review_id | employee_id | review_date | rating |
+-----------+-------------+-------------+--------+
| 1         | 1           | 2023-01-15  | 2      |
| 2         | 1           | 2023-04-15  | 3      |
| 3         | 1           | 2023-07-15  | 4      |
| 4         | 1           | 2023-10-15  | 5      |
| 5         | 2           | 2023-02-01  | 3      |
| 6         | 2           | 2023-05-01  | 2      |
| 7         | 2           | 2023-08-01  | 4      |
| 8         | 2           | 2023-11-01  | 5      |
| 9         | 3           | 2023-03-10  | 1      |
| 10        | 3           | 2023-06-10  | 2      |
| 11        | 3           | 2023-09-10  | 3      |
| 12        | 3           | 2023-12-10  | 4      |
| 13        | 4           | 2023-01-20  | 4      |
| 14        | 4           | 2023-04-20  | 4      |
| 15        | 4           | 2023-07-20  | 4      |
| 16        | 5           | 2023-02-15  | 3      |
| 17        | 5           | 2023-05-15  | 2      |
+-----------+-------------+-------------+--------+

输出：

+-------------+----------------+-------------------+
| employee_id | name           | improvement_score |
+-------------+----------------+-------------------+
| 2           | Bob Smith      | 3                 |
| 1           | Alice Johnson  | 2                 |
| 3           | Carol Davis    | 2                 |
+-------------+----------------+-------------------+

解释：

• Alice Johnson (employee_id = 1)：
  • 有 4 次评估，分数：2, 3, 4, 5
  • 最后 3 次评估（按日期）：2023-04-15 (3), 2023-07-15 (4), 2023-10-15 (5)
  • 评分严格递增：3 → 4 → 5
  • 进步分数：5 - 3 = 2
• Carol Davis (employee_id = 3)：
  • 有 4 次评估，分数：1, 2, 3, 4
  • 最后 3 次评估（按日期）：2023-06-10 (2)，2023-09-10 (3)，2023-12-10 (4)
  • 评分严格递增：2 → 3 → 4
  • 进步分数：4 - 2 = 2
• Bob Smith (employee_id = 2)：
  • 有 4 次评估，分数：3，2，4，5
  • 最后 3 次评估（按日期）：2023-05-01 (2)，2023-08-01 (4)，2023-11-01 (5)
  • 评分严格递增：2 → 4 → 5
  • 进步分数：5 - 2 = 3
• 未包含的员工：
  • David Wilson (employee_id = 4)：之前 3 次评估都是 4 分（没有进步）
  • Emma Brown (employee_id = 5)：只有 2 次评估（需要至少 3 次）

输出表以 improvement_score 降序排序，然后以 name 升序排序。

方法1 （窗口函数 row_number 和 lag，可扩展到前 k 次）

select 
    employee_id,
    name, 
    max(rating) - min(rating) as improvement_score 
from (
    select
        p.employee_id as employee_id,
        e.name as name,
        rating,
        lag(rating, 1, 6) over(partition by p.employee_id order by review_date desc) as pre_rating,
        row_number() over(partition by p.employee_id order by review_date desc) as rn
    from performance_reviews p
    inner join employees e on p.employee_id = e.employee_id
) t1
where rating < pre_rating
and rn <= 3
group by employee_id, name
having count(*) = 3
order by improvement_score desc, name asc

方法2：自连接（不推荐）

with t as (
    select
        *
    from (
        select
            *,
            row_number() over(partition by employee_id order by review_date desc) as rn
        from performance_reviews 
    ) t1
    where rn between 1 and 3
)

select 
    t1.employee_id as employee_id,
    e.name as name, 
    t1.rating - t3.rating as improvement_score 
from t t1
inner join t t2 on t1.employee_id = t2.employee_id and t1.rn = 1 and t2.rn = 2
inner join t t3 on t1.employee_id = t3.employee_id and t1.rn = 1 and t3.rn = 3
inner join employees e on t1.employee_id = e.employee_id
where t1.rating > t2.rating and t2.rating > t3.rating
order by improvement_score desc, name asc