文章目录
- 1 题目描述
- 2 测试用例
- 3 解题思路
- 3.1 解法 1:
1 题目描述
Table: Activity
+--------------+---------+
| Column Name | Type |
+--------------+---------+
| player_id | int |
| device_id | int |
| event_date | date |
| games_played | int |
+--------------+---------+
(player_id, event_date) 是此表的主键 (具有唯一值的列的组合).
这张表显示了某些游戏的玩家的活动情况.
每一行是一个玩家的记录, 他在某一天使用某个设备注销之前登录并玩了很多游戏 (可能是 0).
2 测试用例
编写解决方案, 报告在首次登录的第二天再次登录的玩家的比率, 四舍五入到小数点后两位. 换句话说, 你需要计算从首次登录日期开始至少连续两天登录的玩家的数量, 然后除以玩家总数.
结果格式如下所示:
**示例 1: **
**输入: **
Activity table:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1 | 2 | 2016-03-01 | 5 |
| 1 | 2 | 2016-03-02 | 6 |
| 2 | 3 | 2017-06-25 | 1 |
| 3 | 1 | 2016-03-02 | 0 |
| 3 | 4 | 2018-07-03 | 5 |
+-----------+-----------+------------+--------------+
**输出: **
+-----------+
| fraction |
+-----------+
| 0.33 |
+-----------+
**解释: **
只有 ID 为 1 的玩家在第一天登录后才重新登录, 所以答案是 1/3 = 0.33
3 解题思路
- 先按照
player_id分组统计每个玩家第一次登录的时间min(event_date) as loginDate - 左外链接
Activity查找第一次登录后, 第二天有登录的玩家left join Activity b on a.player_id = b.player_id and datediff(b.event_date, a.loginDate) = 1, 需要用到datediff()函数 - 计算首次登录后连续两天都登录的玩家占所有玩家的比例, 需要用到
avg()函数
3.1 解法 1:
解题思路的步骤 1
select player_id, min(event_date) as loginDate from Activity group by player_id
执行结果
| player_id | loginDate |
|---|---|
| 1 | 2016-03-01 |
| 2 | 2017-06-25 |
| 3 | 2016-03-02 |
解题思路的步骤 2
select a.*, b.*
from (select player_id, min(event_date) as loginDate from Activity group by player_id) as a left join Activity b on a.player_id = b.player_id and datediff(b.event_date, a.loginDate) = 1;
执行结果
| player_id | loginDate | player_id | device_id | event_date | games_played |
|---|---|---|---|---|---|
| 1 | 2016-03-01 | 1 | 2 | 2016-03-02 | 6 |
| 2 | 2017-06-25 | null | null | null | null |
| 3 | 2016-03-02 | null | null | null | null |
解题思路的步骤 3
select round(avg(b.event_date is not null), 2) as fraction
from (select player_id, min(event_date) as loginDate from Activity group by player_id) as a left join Activity b on a.player_id = b.player_id and datediff(b.event_date, a.loginDate) = 1;
执行结果
| fraction |
|---|
| 0.33 |
)
