LeetCode（数据库）- Hopper Company Queries II

Posted 2021-12-20 程序员牧码

题目链接：点击打开链接

题目大意：略。

解题思路：题目英文描述，其实就是比上一题计算 accepted_rides 时，不是全部符合的司机，而是不同的司机，所以需要去重。

估计就是分子这讲述的 at least one，重复的司机选择一个算有效司机即可。

AC 代码

WITH RECURSIVE cal AS(
    SELECT 1 month
    UNION ALL
    SELECT month + 1 FROM cal WHERE month < 12
),

# 计算 active_drivers
t1 AS(SELECT month, COUNT(*) active_drivers
FROM cal LEFT JOIN Drivers ON DATEDIFF(CONCAT('2020-', month, '-1'), DATE_FORMAT(join_date, '%Y-%m-1')) >= 0
WHERE join_date < '2021-1-1'
GROUP BY month),

# 计算 accepted_rides
t2 AS(SELECT month, COUNT(DISTINCT driver_id) accepted_rides
FROM cal LEFT JOIN Drivers ON DATEDIFF(CONCAT('2020-', month, '-1'), DATE_FORMAT(join_date, '%Y-%m-1')) >= 0 
LEFT JOIN (Rides JOIN AcceptedRides USING(ride_id)) USING(driver_id) 
WHERE requested_at BETWEEN '2020-01-01' AND '2020-12-31' AND month = MONTH(requested_at)
GROUP BY month),

# 合并
t3 AS(SELECT month, active_drivers, 0 accepted_rides FROM t1
UNION ALL
SELECT month, 0 active_drivers, accepted_rides FROM t2
UNION ALL
SELECT month, 0 active_drivers, 0 accepted_rides FROM cal)

SELECT month, IFNULL(ROUND(SUM(accepted_rides) / SUM(active_drivers) * 100, 2), 0) working_percentage
FROM t3
GROUP BY month
ORDER BY month