LeetCode（数据库）- 每次访问的交易次数-蒲公英云

LeetCode（数据库）- 每次访问的交易次数

题目链接：点击打开链接

题目大意：略。

解题思路：注意解决方案(1)中，生成序数表可以借助于窗口函数 ROW_NUMBER() OVER()；本题中如果采用解决方案(2)，在第 15 个测试用例中，会出现交易表为空的情况，所以在做表连接的时候处理下笛卡儿积为空的技巧。

AC 代码

-- 解决方案(1)
SELECT *
FROM
(
    SELECT t5.rnb AS transactions_count, IFNULL(visits_count, 0) AS visits_count
    FROM
    (
        SELECT 0 AS rnb
        UNION
        SELECT ROW_NUMBER() OVER () AS rnb
        FROM Transactions
    ) t5
    LEFT JOIN
    (
        SELECT
            cnt AS transactions_count
            ,COUNT(user_id) AS visits_count
        FROM
        (
            SELECT t1.user_id, COUNT(t2.amount) AS cnt
            FROM Visits t1
            LEFT JOIN Transactions t2
            ON t1.user_id = t2.user_id AND t1.visit_date = t2.transaction_date
            GROUP BY user_id, visit_date
        ) t3
        GROUP BY cnt
    ) t4
    ON t5.rnb = t4.transactions_count
) t6
WHERE transactions_count <= (
    SELECT COUNT(t2.amount) AS cnt
    FROM Visits t1
    LEFT JOIN Transactions t2
    ON t1.user_id = t2.user_id AND t1.visit_date = t2.transaction_date
    GROUP BY t1.user_id, visit_date
    ORDER BY cnt DESC
    LIMIT 1)
;
-- 解决方案(2)
# 计算每个用户每日期的交易次数
WITH t1 AS(SELECT user_id, transaction_date, COUNT(*) cnt FROM Transactions GROUP BY user_id, transaction_date),
# 计算每种交易次数对应的用户数(包含交易次数为 0 的情况)
t2 AS(SELECT 0 transactions_count, SUM(IF(t1.cnt IS NULL, 1, 0)) visits_count
FROM Visits v LEFT JOIN t1 ON v.user_id = t1.user_id AND v.Visit_date = t1.transaction_date
UNION ALL
SELECT cnt, COUNT(*) FROM t1 GROUP BY cnt),
# 生成一张最多 Transactions.length 条记录的序数表
t3 AS(SELECT CAST(@i := @i + 1 AS SIGNED) transactions_count, maxn
FROM (SELECT @i:=-1, MAX(transactions_count) maxn FROM t2) init LEFT JOIN Transactions ON 1 = 1)
# 为中途没有的交易次数种类填充
SELECT t3.transactions_count, IFNULL(t2.visits_count, 0) visits_count
FROM t3 LEFT JOIN t2 ON t3.transactions_count = t2.transactions_count
WHERE t3.transactions_count <= maxn