LeetCode-数据库-2

584. 寻找用户推荐人

SELECT
    name
FROM
    Customer
WHERE
    referee_id <> 2
    OR
    referee_id IS NULL

577. 员工奖金

SELECT
    E.name,
    B.bonus
FROM
    Employee E
    LEFT JOIN Bonus B
    ON E.empId = B.empId
WHERE
    B.bonus < 1000
    OR
    B.bonus IS NULL

570. 至少有5名直接下属的经理

SELECT 
    E1.name
FROM 
    Employee E1 
    LEFT JOIN Employee E2
    ON E1.id = E2.managerId
GROUP BY
    E1.id
HAVING
    COUNT(E1.id) >= 5

596. 超过5名学生的课

SELECT
    class
FROM
    Courses
GROUP BY
    class
HAVING
    COUNT(class) >= 5

595. 大的国家

SELECT
    name,
    population,
    area
FROM
    World
WHERE
    population >= 25000000
    OR
    area >= 3000000

586. 订单最多的客户

SELECT
    customer_number
FROM
    Orders
GROUP BY
    customer_number
ORDER BY
    COUNT(customer_number) DESC
LIMIT 1

585. 2016年的投资

思路1

• join两个Insurance表，连接方式为pid不同
• 通过where筛选出2015投资相等的行
• 再来一个子查询找出所有location只出现一次的id
• 通过where筛选出满足条件的id

sql

SELECT
    ROUND(SUM(T.tiv_2016), 2) as tiv_2016
FROM
(
    SELECT
        MAX(I1.tiv_2016) as tiv_2016
    FROM
        Insurance I1
        LEFT JOIN
        Insurance I2
        ON I1.pid != I2.pid
    WHERE 
        I1.pid IN
        (
            SELECT
                pid
            FROM
                Insurance
            GROUP BY
                lat, lon
            HAVING
                COUNT(*) = 1
        )
        AND
            I1.tiv_2015 = I2.tiv_2015
    GROUP BY
        I1.pid
) T

错误分析

• ON和 WHERE的条件可以互换吗？

  SELECT
      ROUND(SUM(T.tiv_2016), 2) as tiv_2016
  FROM
  (
      SELECT
          I1.pid,
          MAX(I1.tiv_2016) as tiv_2016
      FROM
          Insurance I1
          LEFT JOIN
          Insurance I2
          ON I1.pid != I2.pid AND
          I1.tiv_2015 = I2.tiv_2015 AND 
          I1.pid IN
          (
              SELECT
                  pid
              FROM
                  Insurance
              GROUP BY
                  lat, lon
              HAVING
                  COUNT(*) = 1
          )
      GROUP BY
          I1.pid
  ) T

  发现上面的结果不对，原因是使用了LEFT JOIN, 导致On后的条件没有满足，但是左侧都被保留了下来，需要改成内连接

SELECT
    ROUND(SUM(T.tiv_2016), 2) as tiv_2016
FROM
(
    SELECT
        I1.pid,
        MAX(I1.tiv_2016) as tiv_2016
    FROM
        Insurance I1
        JOIN
        Insurance I2
        ON I1.pid != I2.pid AND
        I1.tiv_2015 = I2.tiv_2015 AND 
        I1.pid IN
        (
            SELECT
                pid
            FROM
                Insurance
            GROUP BY
                lat, lon
            HAVING
                COUNT(*) = 1
        )
    GROUP BY
        I1.pid
) T

思路2

• count() + over(parition by)
  • count计算个数
  • partition by指定计算时的聚合方法

sql

SELECT
    ROUND(SUM(T.tiv_2016), 2) as tiv_2016
FROM
(
    SELECT
        tiv_2016,
        COUNT(1) OVER(PARTITION BY tiv_2015) AS tiv_2015_cnt,
        COUNT(1) OVER(PARTITION BY lat, lon) AS pos_cnt
    FROM
        Insurance
) T
WHERE 
    T.tiv_2015_cnt > 1
    AND
    T.pos_cnt = 1

602. 好友申请 II ：谁有最多的好友

思路1

• a向b申请好友，通过后，a-b都互为好友
• 所以需要把requester和accepter互换后，使用UNION ALL连接
• 然后group by + count()

SELECT
    id,
    COUNT(friend_id) AS num
FROM
(
    SELECT
        accepter_id AS id,
        requester_id AS friend_id
    FROM
        RequestAccepted
UNION ALL
    SELECT
        requester_id AS id,
        accepter_id AS friend_id
    FROM
        RequestAccepted
) T
GROUP BY
    id
ORDER BY
    COUNT(friend_id) DESC
LIMIT 1

思路2

• 题目的意思是不存在重复添加好友的情况，比如a加b，a删b，b加回a等情况
• 所以可以直接在子查询中计数，然后在外面求和，可以快一点

SELECT
    id,
    SUM(num) AS num
FROM
(
    SELECT
        accepter_id AS id,
        COUNT(requester_id) AS num
    FROM
        RequestAccepted
    GROUP BY
        accepter_id
UNION ALL
    SELECT
        requester_id AS id,
        COUNT(accepter_id) AS num
    FROM
        RequestAccepted
    GROUP BY
        requester_id
) T
GROUP BY
    id
ORDER BY
    SUM(num) DESC
LIMIT 1

1661. 每台机器的进程平均运行时间

• union前后必须加括号

SELECT
    A1.machine_id,
    ROUND(AVG(A2.timestamp - A1.timestamp), 3) AS processing_time
FROM
    Activity A1
    JOIN
    Activity A2
    ON 
        A1.machine_id = A2.machine_id AND
        A1.process_id = A2.process_id AND
        A1.activity_type = 'start' AND
        A2.activity_type = 'end'
GROUP BY
    A1.machine_id

1341. 电影评分

(
    SELECT 
        name AS results
    FROM
        Users
        JOIN 
            MovieRating
        ON 
            Users.user_id = MovieRating.user_id
    GROUP BY
        Users.user_id
    ORDER BY
        COUNT(rating) DESC,name
    LIMIT 1
)
UNION ALL
(
    SELECT
        title AS results
    FROM
        Movies
        JOIN
            MovieRating
        ON
            Movies.movie_id = MovieRating.movie_id
    WHERE
        created_at BETWEEN DATE("2020-02-01") AND DATE("2020-02-29")
    GROUP BY
        Movies.movie_id
    ORDER BY
        AVG(rating) DESC, title
    LIMIT 1
)

1141. 查询近30天活跃用户数

• 笨蛋，最近30天是往前数29天

SELECT
    activity_date AS 'day',
    COUNT(DISTINCT user_id) AS 'active_users'
FROM
    Activity
WHERE
    activity_date 
    BETWEEN 
        DATE_SUB("2019-07-27", INTERVAL 29 DAY) 
    AND
        DATE("2019-07-27")
GROUP BY
    activity_date

1731. 每位经理的下属员工数量

# Write your MySQL query statement below
SELECT
    E1.employee_id,
    E1.name,
    COUNT(E2.reports_to) AS reports_count,
    ROUND(AVG(E2.age), 0) AS average_age
FROM
    Employees E1
    JOIN
        Employees E2
    ON
        E1.employee_id = E2.reports_to
GROUP BY
    employee_id
ORDER BY
    employee_id

1527. 患某种疾病的患者

LIKE

# Write your MySQL query statement below
SELECT
    patient_id,
    patient_name,
    conditions
FROM
    Patients
WHERE
    conditions LIKE '% DIAB1%'
    OR
    conditions LIKE 'DIAB1%'

REGEXP

• 写不出这个正则
• \b:匹配一个单词边界，也就是指单词和空格间的位置。例如， ‘er\b’ 可以匹配”never” 中的 ‘er’，但不能匹配 “verb” 中的 ‘er’。
• \B:匹配非单词边界。’er\B’ 能匹配 “verb” 中的 ‘er’，但不能匹配 “never” 中的 ‘er’

# Write your MySQL query statement below
SELECT
    patient_id,
    patient_name,
    conditions
FROM
    Patients
WHERE
    conditions REGEXP '\\bDIAB1'

1070. 产品销售分析 III

SELECT
    Sales.product_id,
    T.first_year,
    Sales.quantity,
    Sales.price
FROM
    Sales
    JOIN
    (
        SELECT
            product_id,
            MIN(year) AS first_year
        FROM
            Sales
        GROUP BY
            product_id
    ) T
    ON
        Sales.year = T.first_year
        AND
        Sales.product_id = T.product_id

rank

• over里面既能partition by, 也能order by

SELECT
    T.product_id,
    T.first_year,
    T.quantity,
    T.price
FROM
    (
        SELECT
            product_id,
            year AS first_year,
            quantity,
            price,
            rank() OVER(PARTITION BY product_id ORDER BY year) AS 'year_rank'
        FROM
            Sales
    ) T
WHERE
    T.year_rank = 1