因是個人總結，只列出對自己有用的或較難的：
下面這道題，第一次拿到，我嘗試用 開窗函數 ROW_NUMBER()OVER() 編號，但是發現不能夠處理好連續的問題，
上網查找了別人的解法記錄下來，其實原理 是 把 Logs 看成 三張表，每張表之間關聯 -1 ，然后篩選出 Num 相等的
編寫一個 SQL 查詢，查找所有至少連續出現三次的數字。
+----+-----+
| Id | Num |
+----+-----+
| 1 | 1 |
| 2 | 1 |
| 3 | 1 |
| 4 | 2 |
| 5 | 1 |
| 6 | 2 |
| 7 | 2 |
+----+-----+
例如，給定上面的 Logs 表， 1 是唯一連續出現至少三次的數字。
+-----------------+
| ConsecutiveNums |
+-----------------+
| 1 |
+-----------------+
select distinct log1.Num from Logs log1 join Logs log2 on log1.Id = log2.Id-1
join Logs log3 on log2.Id = log3.Id -1
where log1.Num = log2.Num and log2.Num = log3.Num
-- 同樣的 連續問題
X 市建了一個新的體育館，每日人流量信息被記錄在這三列信息中：序號 (id)、日期 (date)、 人流量 (people)。
請編寫一個查詢語句，找出高峰期時段，要求連續三天及以上，並且每天人流量均不少於100。
例如，表 stadium：
+------+------------+-----------+
| id | date | people |
+------+------------+-----------+
| 1 | 2017-01-01 | 10 |
| 2 | 2017-01-02 | 109 |
| 3 | 2017-01-03 | 150 |
| 4 | 2017-01-04 | 99 |
| 5 | 2017-01-05 | 145 |
| 6 | 2017-01-06 | 1455 |
| 7 | 2017-01-07 | 199 |
| 8 | 2017-01-08 | 188 |
+------+------------+-----------+
對於上面的示例數據，輸出為：
+------+------------+-----------+
| id | date | people |
+------+------------+-----------+
| 5 | 2017-01-05 | 145 |
| 6 | 2017-01-06 | 1455 |
| 7 | 2017-01-07 | 199 |
| 8 | 2017-01-08 | 188 |
+------+------------+-----------+
Note:
每天只有一行記錄，日期隨着 id 的增加而增加。
;WITH stadium(id,date,people) AS
(
SELECT 1,' 2017-01-01 ',10 union all
SELECT 2,' 2017-01-02 ',109 union all
SELECT 3,' 2017-01-03 ',150 union all
SELECT 4,' 2017-01-04 ',99 union all
SELECT 5,' 2017-01-05 ',145 union all
SELECT 6,' 2017-01-06 ',1455 union all
SELECT 7,' 2017-01-07 ',199 union all
SELECT 8,' 2017-01-08 ',188
)
, temp as(
select * from stadium where people > 100 --- 這里 可以避免后面寫一大段 篩選條件
)
,tab as(
select id
,date
,people
,(id - (select max(id) from temp t where t.id < tmp.id)) as id_diff
from temp tmp
)
,final_tab as (
select t2.* from tab t1 join tab t2 on t1.id+1 = t2.id
join tab t3 on t2.id+1 = t3.id
where t1.id_diff = 1 and t2.id_diff = 1 and t3.id_diff = 1
)
select s.* from stadium s join final_tab f on s.id+1 = f.id
union ALL
select id
,date
,people from final_tab
union ALL
select s.* from stadium s join final_tab f on s.id-1 = f.id
leetcode 標記難度 困難
需求：
-- 部門工資前三高的員工
Employee 表包含所有員工信息，每個員工有其對應的 Id, salary 和 department Id 。
+----+-------+--------+--------------+
| Id | Name | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1 | Joe | 70000 | 1 |
| 2 | Henry | 80000 | 2 |
| 3 | Sam | 60000 | 2 |
| 4 | Max | 90000 | 1 |
| 5 | Janet | 69000 | 1 |
| 6 | Randy | 85000 | 1 |
+----+-------+--------+--------------+
Department 表包含公司所有部門的信息。
+----+----------+
| Id | Name |
+----+----------+
| 1 | IT |
| 2 | Sales |
+----+----------+
編寫一個 SQL 查詢，找出每個部門工資前三高的員工。例如，根據上述給定的表格，查詢結果應返回：
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT | Max | 90000 |
| IT | Randy | 85000 |
| IT | Joe | 70000 |
| Sales | Henry | 80000 |
| Sales | Sam | 60000 |
+------------+----------+--------+
第一種解法：
;with Employee(Id,Name,Salary,DepartmentId) AS
(
select 1,'Joe','70000',1 union all
select 2,'Henry','80000',2 union all
select 3,'Sam','60000',2 union all
select 4,'Max','90000',1 union all
select 5,'Janet','69000',1 union all
select 6,'Randy','85000',1
)
, Department(Id,Name) AS(
SELECT 1,'IT'
UNION ALL
SELECT 2,'Sales'
)
select d.Name as Department,e.Name as Employee,e.Salary from
(
SELECT *,ROW_NUMBER()OVER(partition by DepartmentId order by Salary desc) as Rank FROM Employee) e
join Department d on e.DepartmentId = d.Id
where Rank<=3
order by d.Id ASC
第二種解法：
為了避免 有相同排名出現，采用 DENSE_RANK 密級排名
select d.Name as Department,e.Name as Employee,e.Salary from
(
SELECT *,Dense_Rank()OVER(partition by DepartmentId order by Salary desc) as Rank FROM Employee) e
join Department d on e.DepartmentId = d.Id
where Rank<=3
order by d.Id ASC
leetcode 難度標記： 簡單
給定一個 Weather 表，編寫一個 SQL 查詢，來查找與之前（昨天的）日期相比溫度更高的所有日期的 Id。
+---------+------------------+------------------+
| Id(INT) | RecordDate(DATE) | Temperature(INT) |
+---------+------------------+------------------+
| 1 | 2015-01-01 | 10 |
| 2 | 2015-01-02 | 25 |
| 3 | 2015-01-03 | 20 |
| 4 | 2015-01-04 | 30 |
+---------+------------------+------------------+
例如，根據上述給定的 Weather 表格，返回如下 Id:
+----+
| Id |
+----+
| 2 |
| 4 |
+----+
select Id from Weather w where Temperature > (select Temperature from Weather l where dateadd(day,1,l.RecordDate) = w.RecordDate)
leetcode 難度標記： 簡單
編寫一個 SQL 查詢，獲取 Employee 表中第二高的薪水（Salary） 。
+----+--------+
| Id | Salary |
+----+--------+
| 1 | 100 |
| 2 | 200 |
| 3 | 300 |
+----+--------+
例如上述 Employee 表，SQL查詢應該返回 200 作為第二高的薪水。如果不存在第二高的薪水，那么查詢應返回 null。
+---------------------+
| SecondHighestSalary |
+---------------------+
| 200 |
+---------------------+
解法：
SELECT MAX(SALARY) AS SecondHighestSalary FROM Employee WHERE Salary <(SELECT MAX(SALARY) FROM Employee)
leetcode 難度標記： 中等
Employee 表包含所有員工信息，每個員工有其對應的 Id, salary 和 department Id。
+----+-------+--------+--------------+
| Id | Name | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1 | Joe | 70000 | 1 |
| 2 | Henry | 80000 | 2 |
| 3 | Sam | 60000 | 2 |
| 4 | Max | 90000 | 1 |
+----+-------+--------+--------------+
Department 表包含公司所有部門的信息。
+----+----------+
| Id | Name |
+----+----------+
| 1 | IT |
| 2 | Sales |
+----+----------+
編寫一個 SQL 查詢，找出每個部門工資最高的員工。例如，根據上述給定的表格，Max 在 IT 部門有最高工資，Henry 在 Sales 部門有最高工資。
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT | Max | 90000 |
| Sales | Henry | 80000 |
+------------+----------+--------+
/* Write your T-SQL query statement below */
select d.Name as Department
,e.Name as Employee
,e.Salary as Salary
from Employee e
join Department d on e.DepartmentId = d.Id
join( SELECT DepartmentId
,max(Salary) as Salary
FROM Employee Group by DepartmentId
)
grp on e.Salary = grp.Salary
and e.DepartmentId = grp.DepartmentId
編寫一個 SQL 查詢，來刪除 Person 表中所有重復的電子郵箱，重復的郵箱里只保留 Id 最小 的那個。
+----+------------------+
| Id | Email |
+----+------------------+
| 1 | john@example.com |
| 2 | bob@example.com |
| 3 | john@example.com |
+----+------------------+
Id 是這個表的主鍵。
例如，在運行你的查詢語句之后，上面的 Person 表應返回以下幾行:
+----+------------------+
| Id | Email |
+----+------------------+
| 1 | john@example.com |
| 2 | bob@example.com |
+----+------------------+
--- 這道題 leetcode 只支持 mysql 語法，很是惡心到我了。。。最后也不知道 運行成功了沒有，但是 暫且記下 我的答案吧
# Write your MySQL query statement below
delete from Person where id not in(select id from ( select min(Id) as id
from Person
where Email in (select Email
from Person
group by Email
having count(Email) > 1
) ) as n )
and email in (select email from (select Email
from Person
group by Email
having count(Email)>1) as m)
leetcode 難度標記：中等
小美是一所中學的信息科技老師，她有一張 seat 座位表，平時用來儲存學生名字和與他們相對應的座位 id。
其中縱列的 id 是連續遞增的
小美想改變相鄰倆學生的座位。
你能不能幫她寫一個 SQL query 來輸出小美想要的結果呢？
示例：
+---------+---------+
| id | student |
+---------+---------+
| 1 | Abbot |
| 2 | Doris |
| 3 | Emerson |
| 4 | Green |
| 5 | Jeames |
+---------+---------+
假如數據輸入的是上表，則輸出結果如下：
+---------+---------+
| id | student |
+---------+---------+
| 1 | Doris |
| 2 | Abbot |
| 3 | Green |
| 4 | Emerson |
| 5 | Jeames |
+---------+---------+
注意：
如果學生人數是奇數，則不需要改變最后一個同學的座位。
;with seat(id,student) as(
SELECT 1 ,'Abbot' union all
SELECT 2 ,'Doris' union all
SELECT 3 ,'Emerson' union all
SELECT 4 ,'Green' union all
SELECT 5 ,'Jeames'
)
,odd_tab as(
select * from seat where id %2 = 1
)
,even_tab as(
select * from seat where id %2 = 0
)
select o.id,isnull(e.student,o.student) as student from odd_tab o left join even_tab e on o.id +1 = e.id
union ALL
select e.id,o.student from odd_tab o join even_tab e on o.id +1 = e.id
order by id asc
leetcode 難度標記 困難，個人覺得不該標記為困難
Trips 表中存所有出租車的行程信息。每段行程有唯一健 Id，Client_Id 和 Driver_Id 是 Users 表中 Users_Id 的外鍵。Status 是枚舉類型，枚舉成員為 (‘completed’, ‘cancelled_by_driver’, ‘cancelled_by_client’)。
+----+-----------+-----------+---------+--------------------+----------+
| Id | Client_Id | Driver_Id | City_Id | Status |Request_at|
+----+-----------+-----------+---------+--------------------+----------+
| 1 | 1 | 10 | 1 | completed |2013-10-01|
| 2 | 2 | 11 | 1 | cancelled_by_driver|2013-10-01|
| 3 | 3 | 12 | 6 | completed |2013-10-01|
| 4 | 4 | 13 | 6 | cancelled_by_client|2013-10-01|
| 5 | 1 | 10 | 1 | completed |2013-10-02|
| 6 | 2 | 11 | 6 | completed |2013-10-02|
| 7 | 3 | 12 | 6 | completed |2013-10-02|
| 8 | 2 | 12 | 12 | completed |2013-10-03|
| 9 | 3 | 10 | 12 | completed |2013-10-03|
| 10 | 4 | 13 | 12 | cancelled_by_driver|2013-10-03|
+----+-----------+-----------+---------+--------------------+----------+
Users 表存所有用戶。每個用戶有唯一鍵 Users_Id。Banned 表示這個用戶是否被禁止，Role 則是一個表示（‘client’, ‘driver’, ‘partner’）的枚舉類型。
+----------+--------+--------+
| Users_Id | Banned | Role |
+----------+--------+--------+
| 1 | No | client |
| 2 | Yes | client |
| 3 | No | client |
| 4 | No | client |
| 10 | No | driver |
| 11 | No | driver |
| 12 | No | driver |
| 13 | No | driver |
+----------+--------+--------+
寫一段 SQL 語句查出 2013年10月1日 至 2013年10月3日 期間非禁止用戶的取消率。基於上表，你的 SQL 語句應返回如下結果，取消率（Cancellation Rate）保留兩位小數。
+------------+-------------------+
| Day | Cancellation Rate |
+------------+-------------------+
| 2013-10-01 | 0.33 |
| 2013-10-02 | 0.00 |
| 2013-10-03 | 0.50 |
+------------+-------------------+
Create table Trips (Id int,Client_Id int,
Driver_Id int, City_Id int,
Status varchar(50),
Request_at varchar(50));
Create table Users (Users_Id int,Banned varchar(50), Role varchar(50));
Truncate table Trips;
insert into Trips (Id, Client_Id, Driver_Id,City_Id, Status, Request_at) values ('1', '1', '10', '1', 'completed','2013-10-01');
insert into Trips (Id, Client_Id, Driver_Id,City_Id, Status, Request_at) values ('2', '2', '11', '1','cancelled_by_driver', '2013-10-01');
insert into Trips (Id, Client_Id, Driver_Id,City_Id, Status, Request_at) values ('3', '3', '12', '6', 'completed','2013-10-01');
insert into Trips (Id, Client_Id, Driver_Id,City_Id, Status, Request_at) values ('4', '4', '13', '6','cancelled_by_client', '2013-10-01');
insert into Trips (Id, Client_Id, Driver_Id,City_Id, Status, Request_at) values ('5', '1', '10', '1', 'completed','2013-10-02');
insert into Trips (Id, Client_Id, Driver_Id,City_Id, Status, Request_at) values ('6', '2', '11', '6', 'completed','2013-10-02');
insert into Trips (Id, Client_Id, Driver_Id,City_Id, Status, Request_at) values ('7', '3', '12', '6', 'completed','2013-10-02');
insert into Trips (Id, Client_Id, Driver_Id,City_Id, Status, Request_at) values ('8', '2', '12', '12', 'completed','2013-10-03');
insert into Trips (Id, Client_Id, Driver_Id,City_Id, Status, Request_at) values ('9', '3', '10', '12', 'completed','2013-10-03');
insert into Trips (Id, Client_Id, Driver_Id,City_Id, Status, Request_at) values ('10', '4', '13', '12','cancelled_by_driver', '2013-10-03');
Truncate table Users;
insert into Users (Users_Id, Banned, Role)values ('1', 'No', 'client');
insert into Users (Users_Id, Banned, Role)values ('2', 'Yes', 'client');
insert into Users (Users_Id, Banned, Role)values ('3', 'No', 'client');
insert into Users (Users_Id, Banned, Role)values ('4', 'No', 'client');
insert into Users (Users_Id, Banned, Role)values ('10', 'No', 'driver');
insert into Users (Users_Id, Banned, Role)values ('11', 'No', 'driver');
insert into Users (Users_Id, Banned, Role)values ('12', 'No', 'driver');
insert into Users (Users_Id, Banned, Role)values ('13', 'No', 'driver');
select t.Request_at as [Day],Round((sum(case when Status like '%cancelled%' then 1 else 0 end) * 1.0 / count(status) ),2) as [Cancellation Rate]
from Trips t join USERs u
on t.client_id = u.users_id
where Request_at between '2013-10-01' and '2013-10-03'
and u.Role = 'client' and Banned = 'No'
group by Request_at
小結：
目前，leetcode 上的 sql 題均已刷完，感覺難度一般，個人卡殼的地方 是 連續問題的 求解， 里邊有兩道 連續問題，本文第一道和第二道
下次見到 連續問題應該首先想 多表 id+1 ， id+2 ,...,id+n 這種方式求解，而不應想編號，編號不足以解決連續問題