From 3b3fac2878c0c71c6249b8ef64013d77c9a71064 Mon Sep 17 00:00:00 2001
From: CyC2018 <zhengyc101@163.com>
Date: Wed, 11 Dec 2019 00:19:49 +0800
Subject: [PATCH] auto commit

---
 ...code-Database \351\242\230\350\247\243.md" | 423 +++++++++---------
 ...code-Database \351\242\230\350\247\243.md" | 423 +++++++++---------
 2 files changed, 430 insertions(+), 416 deletions(-)

diff --git "a/docs/notes/Leetcode-Database \351\242\230\350\247\243.md" "b/docs/notes/Leetcode-Database \351\242\230\350\247\243.md"
index 3531c2bb6f..64aed608da 100644
--- "a/docs/notes/Leetcode-Database \351\242\230\350\247\243.md"	
+++ "b/docs/notes/Leetcode-Database \351\242\230\350\247\243.md"	
@@ -46,9 +46,22 @@ https://leetcode.com/problems/big-countries/description/
 +--------------+-------------+--------------+
 ```
 
+## Solution
+
+```sql
+SELECT name,
+    population,
+    area
+FROM
+    World
+WHERE
+    area > 3000000
+    OR population > 25000000;
+```
+
 ## SQL Schema
 
-SQL Schema 用于在本地环境下创建表结构并导入数据，从而方便在本地环境解答。
+SQL Schema 用于在本地环境下创建表结构并导入数据，从而方便在本地环境调试。
 
 ```sql
 DROP TABLE
@@ -64,19 +77,6 @@ VALUES
     ( 'Angola', 'Africa', '1246700', '20609294', '1009900000' );
 ```
 
-## Solution
-
-```sql
-SELECT name,
-    population,
-    area
-FROM
-    World
-WHERE
-    area > 3000000
-    OR population > 25000000;
-```
-
 # 627. Swap Salary
 
 https://leetcode.com/problems/swap-salary/description/
@@ -103,6 +103,24 @@ https://leetcode.com/problems/swap-salary/description/
 | 4  | D    | m   | 500    |
 ```
 
+## Solution
+
+两个相等的数异或的结果为 0，而 0 与任何一个数异或的结果为这个数。
+
+sex  字段只有两个取值：'f' 和 'm'，并且有以下规律：
+
+```
+'f' ^ ('m' ^ 'f') = 'm' ^ ('f' ^ 'f') = 'm'
+'m' ^ ('m' ^ 'f') = 'f' ^ ('m' ^ 'm') = 'f'
+```
+
+因此将 sex 字段和 'm' ^ 'f' 进行异或操作，最后就能反转 sex 字段。
+
+```sql
+UPDATE salary
+SET sex = CHAR ( ASCII(sex) ^ ASCII( 'm' ) ^ ASCII( 'f' ) );
+```
+
 ## SQL Schema
 
 ```sql
@@ -118,22 +136,6 @@ VALUES
     ( '4', 'D', 'f', '500' );
 ```
 
-## Solution
-
-使用异或操作，两个相等的数异或的结果为 0，而 0 与任何一个数异或的结果为这个数。
-
-```
-'f' ^ 'm' ^ 'f' = 'm'
-'m' ^ 'm' ^ 'f' = 'f'
-```
-
-
-
-```sql
-UPDATE salary
-SET sex = CHAR ( ASCII(sex) ^ ASCII( 'm' ) ^ ASCII( 'f' ) );
-```
-
 # 620. Not Boring Movies
 
 https://leetcode.com/problems/not-boring-movies/description/
@@ -164,6 +166,20 @@ https://leetcode.com/problems/not-boring-movies/description/
 +---------+-----------+--------------+-----------+
 ```
 
+## Solution
+
+```sql
+SELECT
+    *
+FROM
+    cinema
+WHERE
+    id % 2 = 1
+    AND description != 'boring'
+ORDER BY
+    rating DESC;
+```
+
 ## SQL Schema
 
 ```sql
@@ -180,20 +196,6 @@ VALUES
     ( 5, 'House card', 'Interesting', 9.1 );
 ```
 
-## Solution
-
-```sql
-SELECT
-    *
-FROM
-    cinema
-WHERE
-    id % 2 = 1
-    AND description != 'boring'
-ORDER BY
-    rating DESC;
-```
-
 # 596. Classes More Than 5 Students
 
 https://leetcode.com/problems/classes-more-than-5-students/description/
@@ -226,6 +228,21 @@ https://leetcode.com/problems/classes-more-than-5-students/description/
 +---------+
 ```
 
+## Solution
+
+对 class 列进行分组之后，再使用 count 汇总函数统计数量，统计之后使用 having 进行过滤。
+
+```sql
+SELECT
+    class
+FROM
+    courses
+GROUP BY
+    class
+HAVING
+    count( DISTINCT student ) >= 5;
+```
+
 ## SQL Schema
 
 ```sql
@@ -246,21 +263,6 @@ VALUES
     ( 'I', 'Math' );
 ```
 
-## Solution
-
-对 class 列进行分组之后，再使用 count 汇总函数统计数量，统计之后使用 having 进行过滤。
-
-```sql
-SELECT
-    class
-FROM
-    courses
-GROUP BY
-    class
-HAVING
-    count( DISTINCT student ) >= 5;
-```
-
 # 182. Duplicate Emails
 
 https://leetcode.com/problems/duplicate-emails/description/
@@ -289,20 +291,6 @@ https://leetcode.com/problems/duplicate-emails/description/
 +---------+
 ```
 
-## SQL Schema
-
-```sql
-DROP TABLE
-IF
-    EXISTS Person;
-CREATE TABLE Person ( Id INT, Email VARCHAR ( 255 ) );
-INSERT INTO Person ( Id, Email )
-VALUES
-    ( 1, 'a@b.com' ),
-    ( 2, 'c@d.com' ),
-    ( 3, 'a@b.com' );
-```
-
 ## Solution
 
 对 Email 进行分组，如果相同 Email 的数量大于等于 2，则表示该 Email 重复。
@@ -318,6 +306,21 @@ HAVING
     COUNT( * ) >= 2;
 ```
 
+## SQL Schema
+
+```sql
+DROP TABLE
+IF
+    EXISTS Person;
+CREATE TABLE Person ( Id INT, Email VARCHAR ( 255 ) );
+INSERT INTO Person ( Id, Email )
+VALUES
+    ( 1, 'a@b.com' ),
+    ( 2, 'c@d.com' ),
+    ( 3, 'a@b.com' );
+```
+
+
 # 196. Delete Duplicate Emails
 
 https://leetcode.com/problems/delete-duplicate-emails/description/
@@ -347,10 +350,6 @@ https://leetcode.com/problems/delete-duplicate-emails/description/
 +----+------------------+
 ```
 
-## SQL Schema
-
-与 182 相同。
-
 ## Solution
 
 只保留相同 Email 中 Id 最小的那一个，然后删除其它的。
@@ -389,6 +388,12 @@ WHERE
 
 参考：[pMySQL Error 1093 - Can't specify target table for update in FROM clause](https://stackoverflow.com/questions/45494/mysql-error-1093-cant-specify-target-table-for-update-in-from-clause)
 
+## SQL Schema
+
+与 182 相同。
+
+
+
 # 175. Combine Two Tables
 
 https://leetcode.com/problems/combine-two-tables/description/
@@ -424,6 +429,22 @@ AddressId is the primary key column for this table.
 
 查找 FirstName, LastName, City, State 数据，而不管一个用户有没有填地址信息。
 
+## Solution
+
+涉及到 Person 和 Address 两个表，在对这两个表执行连接操作时，因为要保留 Person 表中的信息，即使在 Address 表中没有关联的信息也要保留。此时可以用左外连接，将 Person 表放在 LEFT JOIN 的左边。
+
+```sql
+SELECT
+    FirstName,
+    LastName,
+    City,
+    State
+FROM
+    Person P
+    LEFT JOIN Address A
+    ON P.PersonId = A.PersonId;
+```
+
 ## SQL Schema
 
 ```sql
@@ -443,22 +464,6 @@ VALUES
     ( 1, 2, 'New York City', 'New York' );
 ```
 
-## Solution
-
-涉及到 Person 和 Address 两个表，在对这两个表执行连接操作时，因为要保留 Person 表中的信息，即使在 Address 表中没有关联的信息也要保留。此时可以用左外连接，将 Person 表放在 LEFT JOIN 的左边。
-
-```sql
-SELECT
-    FirstName,
-    LastName,
-    City,
-    State
-FROM
-    Person P
-    LEFT JOIN Address A
-    ON P.PersonId = A.PersonId;
-```
-
 # 181. Employees Earning More Than Their Managers
 
 https://leetcode.com/problems/employees-earning-more-than-their-managers/description/
@@ -480,6 +485,18 @@ Employee 表：
 
 查找薪资大于其经理薪资的员工信息。
 
+## Solution
+
+```sql
+SELECT
+    E1.NAME AS Employee
+FROM
+    Employee E1
+    INNER JOIN Employee E2
+    ON E1.ManagerId = E2.Id
+    AND E1.Salary > E2.Salary;
+```
+
 ## SQL Schema
 
 ```sql
@@ -495,18 +512,6 @@ VALUES
     ( 4, 'Max', 90000, NULL );
 ```
 
-## Solution
-
-```sql
-SELECT
-    E1.NAME AS Employee
-FROM
-    Employee E1
-    INNER JOIN Employee E2
-    ON E1.ManagerId = E2.Id
-    AND E1.Salary > E2.Salary;
-```
-
 # 183. Customers Who Never Order
 
 https://leetcode.com/problems/customers-who-never-order/description/
@@ -548,29 +553,6 @@ Orders 表：
 +-----------+
 ```
 
-## SQL Schema
-
-```sql
-DROP TABLE
-IF
-    EXISTS Customers;
-CREATE TABLE Customers ( Id INT, NAME VARCHAR ( 255 ) );
-DROP TABLE
-IF
-    EXISTS Orders;
-CREATE TABLE Orders ( Id INT, CustomerId INT );
-INSERT INTO Customers ( Id, NAME )
-VALUES
-    ( 1, 'Joe' ),
-    ( 2, 'Henry' ),
-    ( 3, 'Sam' ),
-    ( 4, 'Max' );
-INSERT INTO Orders ( Id, CustomerId )
-VALUES
-    ( 1, 3 ),
-    ( 2, 1 );
-```
-
 ## Solution
 
 左外链接
@@ -597,6 +579,29 @@ WHERE
     Id NOT IN ( SELECT CustomerId FROM Orders );
 ```
 
+## SQL Schema
+
+```sql
+DROP TABLE
+IF
+    EXISTS Customers;
+CREATE TABLE Customers ( Id INT, NAME VARCHAR ( 255 ) );
+DROP TABLE
+IF
+    EXISTS Orders;
+CREATE TABLE Orders ( Id INT, CustomerId INT );
+INSERT INTO Customers ( Id, NAME )
+VALUES
+    ( 1, 'Joe' ),
+    ( 2, 'Henry' ),
+    ( 3, 'Sam' ),
+    ( 4, 'Max' );
+INSERT INTO Orders ( Id, CustomerId )
+VALUES
+    ( 1, 3 ),
+    ( 2, 1 );
+```
+
 # 184. Department Highest Salary
 
 https://leetcode.com/problems/department-highest-salary/description/
@@ -638,25 +643,6 @@ Department 表：
 +------------+----------+--------+
 ```
 
-## SQL Schema
-
-```sql
-DROP TABLE IF EXISTS Employee;
-CREATE TABLE Employee ( Id INT, NAME VARCHAR ( 255 ), Salary INT, DepartmentId INT );
-DROP TABLE IF EXISTS Department;
-CREATE TABLE Department ( Id INT, NAME VARCHAR ( 255 ) );
-INSERT INTO Employee ( Id, NAME, Salary, DepartmentId )
-VALUES
-    ( 1, 'Joe', 70000, 1 ),
-    ( 2, 'Henry', 80000, 2 ),
-    ( 3, 'Sam', 60000, 2 ),
-    ( 4, 'Max', 90000, 1 );
-INSERT INTO Department ( Id, NAME )
-VALUES
-    ( 1, 'IT' ),
-    ( 2, 'Sales' );
-```
-
 ## Solution
 
 创建一个临时表，包含了部门员工的最大薪资。可以对部门进行分组，然后使用 MAX() 汇总函数取得最大薪资。
@@ -678,6 +664,26 @@ WHERE
     AND E.Salary = M.Salary;
 ```
 
+## SQL Schema
+
+```sql
+DROP TABLE IF EXISTS Employee;
+CREATE TABLE Employee ( Id INT, NAME VARCHAR ( 255 ), Salary INT, DepartmentId INT );
+DROP TABLE IF EXISTS Department;
+CREATE TABLE Department ( Id INT, NAME VARCHAR ( 255 ) );
+INSERT INTO Employee ( Id, NAME, Salary, DepartmentId )
+VALUES
+    ( 1, 'Joe', 70000, 1 ),
+    ( 2, 'Henry', 80000, 2 ),
+    ( 3, 'Sam', 60000, 2 ),
+    ( 4, 'Max', 90000, 1 );
+INSERT INTO Department ( Id, NAME )
+VALUES
+    ( 1, 'IT' ),
+    ( 2, 'Sales' );
+```
+
+
 # 176. Second Highest Salary
 
 https://leetcode.com/problems/second-highest-salary/description/
@@ -706,6 +712,15 @@ https://leetcode.com/problems/second-highest-salary/description/
 
 没有找到返回 null 而不是不返回数据。
 
+## Solution
+
+为了在没有查找到数据时返回 null，需要在查询结果外面再套一层 SELECT。
+
+```sql
+SELECT
+    ( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT 1, 1 ) SecondHighestSalary;
+```
+
 ## SQL Schema
 
 ```sql
@@ -720,25 +735,12 @@ VALUES
     ( 3, 300 );
 ```
 
-## Solution
-
-为了在没有查找到数据时返回 null，需要在查询结果外面再套一层 SELECT。
-
-```sql
-SELECT
-    ( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT 1, 1 ) SecondHighestSalary;
-```
-
 # 177. Nth Highest Salary
 
 ## Description
 
 查找工资第 N 高的员工。
 
-## SQL Schema
-
-同 176。
-
 ## Solution
 
 ```sql
@@ -750,6 +752,11 @@ RETURN ( SELECT ( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMI
 END
 ```
 
+## SQL Schema
+
+同 176。
+
+
 # 178. Rank Scores
 
 https://leetcode.com/problems/rank-scores/description/
@@ -786,23 +793,6 @@ https://leetcode.com/problems/rank-scores/description/
 +-------+------+
 ```
 
-## SQL Schema
-
-```sql
-DROP TABLE
-IF
-    EXISTS Scores;
-CREATE TABLE Scores ( Id INT, Score DECIMAL ( 3, 2 ) );
-INSERT INTO Scores ( Id, Score )
-VALUES
-    ( 1, 3.5 ),
-    ( 2, 3.65 ),
-    ( 3, 4.0 ),
-    ( 4, 3.85 ),
-    ( 5, 4.0 ),
-    ( 6, 3.65 );
-```
-
 ## Solution
 
 要统计某个 score 的排名，只要统计大于该 score 的 score 数量，然后加 1。
@@ -852,6 +842,23 @@ ORDER BY
     S1.score DESC;
 ```
 
+## SQL Schema
+
+```sql
+DROP TABLE
+IF
+    EXISTS Scores;
+CREATE TABLE Scores ( Id INT, Score DECIMAL ( 3, 2 ) );
+INSERT INTO Scores ( Id, Score )
+VALUES
+    ( 1, 3.5 ),
+    ( 2, 3.65 ),
+    ( 3, 4.0 ),
+    ( 4, 3.85 ),
+    ( 5, 4.0 ),
+    ( 6, 3.65 );
+```
+
 # 180. Consecutive Numbers
 
 https://leetcode.com/problems/consecutive-numbers/description/
@@ -884,6 +891,21 @@ https://leetcode.com/problems/consecutive-numbers/description/
 +-----------------+
 ```
 
+## Solution
+
+```sql
+SELECT
+    DISTINCT L1.num ConsecutiveNums
+FROM
+    Logs L1,
+    Logs L2,
+    Logs L3
+WHERE L1.id = l2.id - 1
+    AND L2.id = L3.id - 1
+    AND L1.num = L2.num
+    AND l2.num = l3.num;
+```
+
 ## SQL Schema
 
 ```sql
@@ -902,21 +924,6 @@ VALUES
     ( 7, 2 );
 ```
 
-## Solution
-
-```sql
-SELECT
-    DISTINCT L1.num ConsecutiveNums
-FROM
-    Logs L1,
-    Logs L2,
-    Logs L3
-WHERE L1.id = l2.id - 1
-    AND L2.id = L3.id - 1
-    AND L1.num = L2.num
-    AND l2.num = l3.num;
-```
-
 # 626. Exchange Seats
 
 https://leetcode.com/problems/exchange-seats/description/
@@ -951,22 +958,6 @@ seat 表存储着座位对应的学生。
 +---------+---------+
 ```
 
-## SQL Schema
-
-```sql
-DROP TABLE
-IF
-    EXISTS seat;
-CREATE TABLE seat ( id INT, student VARCHAR ( 255 ) );
-INSERT INTO seat ( id, student )
-VALUES
-    ( '1', 'Abbot' ),
-    ( '2', 'Doris' ),
-    ( '3', 'Emerson' ),
-    ( '4', 'Green' ),
-    ( '5', 'Jeames' );
-```
-
 ## Solution
 
 使用多个 union。
@@ -1004,6 +995,22 @@ ORDER BY
     id;
 ```
 
+## SQL Schema
+
+```sql
+DROP TABLE
+IF
+    EXISTS seat;
+CREATE TABLE seat ( id INT, student VARCHAR ( 255 ) );
+INSERT INTO seat ( id, student )
+VALUES
+    ( '1', 'Abbot' ),
+    ( '2', 'Doris' ),
+    ( '3', 'Emerson' ),
+    ( '4', 'Green' ),
+    ( '5', 'Jeames' );
+```
+
 
 
 
diff --git "a/notes/Leetcode-Database \351\242\230\350\247\243.md" "b/notes/Leetcode-Database \351\242\230\350\247\243.md"
index 3531c2bb6f..64aed608da 100644
--- "a/notes/Leetcode-Database \351\242\230\350\247\243.md"	
+++ "b/notes/Leetcode-Database \351\242\230\350\247\243.md"	
@@ -46,9 +46,22 @@ https://leetcode.com/problems/big-countries/description/
 +--------------+-------------+--------------+
 ```
 
+## Solution
+
+```sql
+SELECT name,
+    population,
+    area
+FROM
+    World
+WHERE
+    area > 3000000
+    OR population > 25000000;
+```
+
 ## SQL Schema
 
-SQL Schema 用于在本地环境下创建表结构并导入数据，从而方便在本地环境解答。
+SQL Schema 用于在本地环境下创建表结构并导入数据，从而方便在本地环境调试。
 
 ```sql
 DROP TABLE
@@ -64,19 +77,6 @@ VALUES
     ( 'Angola', 'Africa', '1246700', '20609294', '1009900000' );
 ```
 
-## Solution
-
-```sql
-SELECT name,
-    population,
-    area
-FROM
-    World
-WHERE
-    area > 3000000
-    OR population > 25000000;
-```
-
 # 627. Swap Salary
 
 https://leetcode.com/problems/swap-salary/description/
@@ -103,6 +103,24 @@ https://leetcode.com/problems/swap-salary/description/
 | 4  | D    | m   | 500    |
 ```
 
+## Solution
+
+两个相等的数异或的结果为 0，而 0 与任何一个数异或的结果为这个数。
+
+sex  字段只有两个取值：'f' 和 'm'，并且有以下规律：
+
+```
+'f' ^ ('m' ^ 'f') = 'm' ^ ('f' ^ 'f') = 'm'
+'m' ^ ('m' ^ 'f') = 'f' ^ ('m' ^ 'm') = 'f'
+```
+
+因此将 sex 字段和 'm' ^ 'f' 进行异或操作，最后就能反转 sex 字段。
+
+```sql
+UPDATE salary
+SET sex = CHAR ( ASCII(sex) ^ ASCII( 'm' ) ^ ASCII( 'f' ) );
+```
+
 ## SQL Schema
 
 ```sql
@@ -118,22 +136,6 @@ VALUES
     ( '4', 'D', 'f', '500' );
 ```
 
-## Solution
-
-使用异或操作，两个相等的数异或的结果为 0，而 0 与任何一个数异或的结果为这个数。
-
-```
-'f' ^ 'm' ^ 'f' = 'm'
-'m' ^ 'm' ^ 'f' = 'f'
-```
-
-
-
-```sql
-UPDATE salary
-SET sex = CHAR ( ASCII(sex) ^ ASCII( 'm' ) ^ ASCII( 'f' ) );
-```
-
 # 620. Not Boring Movies
 
 https://leetcode.com/problems/not-boring-movies/description/
@@ -164,6 +166,20 @@ https://leetcode.com/problems/not-boring-movies/description/
 +---------+-----------+--------------+-----------+
 ```
 
+## Solution
+
+```sql
+SELECT
+    *
+FROM
+    cinema
+WHERE
+    id % 2 = 1
+    AND description != 'boring'
+ORDER BY
+    rating DESC;
+```
+
 ## SQL Schema
 
 ```sql
@@ -180,20 +196,6 @@ VALUES
     ( 5, 'House card', 'Interesting', 9.1 );
 ```
 
-## Solution
-
-```sql
-SELECT
-    *
-FROM
-    cinema
-WHERE
-    id % 2 = 1
-    AND description != 'boring'
-ORDER BY
-    rating DESC;
-```
-
 # 596. Classes More Than 5 Students
 
 https://leetcode.com/problems/classes-more-than-5-students/description/
@@ -226,6 +228,21 @@ https://leetcode.com/problems/classes-more-than-5-students/description/
 +---------+
 ```
 
+## Solution
+
+对 class 列进行分组之后，再使用 count 汇总函数统计数量，统计之后使用 having 进行过滤。
+
+```sql
+SELECT
+    class
+FROM
+    courses
+GROUP BY
+    class
+HAVING
+    count( DISTINCT student ) >= 5;
+```
+
 ## SQL Schema
 
 ```sql
@@ -246,21 +263,6 @@ VALUES
     ( 'I', 'Math' );
 ```
 
-## Solution
-
-对 class 列进行分组之后，再使用 count 汇总函数统计数量，统计之后使用 having 进行过滤。
-
-```sql
-SELECT
-    class
-FROM
-    courses
-GROUP BY
-    class
-HAVING
-    count( DISTINCT student ) >= 5;
-```
-
 # 182. Duplicate Emails
 
 https://leetcode.com/problems/duplicate-emails/description/
@@ -289,20 +291,6 @@ https://leetcode.com/problems/duplicate-emails/description/
 +---------+
 ```
 
-## SQL Schema
-
-```sql
-DROP TABLE
-IF
-    EXISTS Person;
-CREATE TABLE Person ( Id INT, Email VARCHAR ( 255 ) );
-INSERT INTO Person ( Id, Email )
-VALUES
-    ( 1, 'a@b.com' ),
-    ( 2, 'c@d.com' ),
-    ( 3, 'a@b.com' );
-```
-
 ## Solution
 
 对 Email 进行分组，如果相同 Email 的数量大于等于 2，则表示该 Email 重复。
@@ -318,6 +306,21 @@ HAVING
     COUNT( * ) >= 2;
 ```
 
+## SQL Schema
+
+```sql
+DROP TABLE
+IF
+    EXISTS Person;
+CREATE TABLE Person ( Id INT, Email VARCHAR ( 255 ) );
+INSERT INTO Person ( Id, Email )
+VALUES
+    ( 1, 'a@b.com' ),
+    ( 2, 'c@d.com' ),
+    ( 3, 'a@b.com' );
+```
+
+
 # 196. Delete Duplicate Emails
 
 https://leetcode.com/problems/delete-duplicate-emails/description/
@@ -347,10 +350,6 @@ https://leetcode.com/problems/delete-duplicate-emails/description/
 +----+------------------+
 ```
 
-## SQL Schema
-
-与 182 相同。
-
 ## Solution
 
 只保留相同 Email 中 Id 最小的那一个，然后删除其它的。
@@ -389,6 +388,12 @@ WHERE
 
 参考：[pMySQL Error 1093 - Can't specify target table for update in FROM clause](https://stackoverflow.com/questions/45494/mysql-error-1093-cant-specify-target-table-for-update-in-from-clause)
 
+## SQL Schema
+
+与 182 相同。
+
+
+
 # 175. Combine Two Tables
 
 https://leetcode.com/problems/combine-two-tables/description/
@@ -424,6 +429,22 @@ AddressId is the primary key column for this table.
 
 查找 FirstName, LastName, City, State 数据，而不管一个用户有没有填地址信息。
 
+## Solution
+
+涉及到 Person 和 Address 两个表，在对这两个表执行连接操作时，因为要保留 Person 表中的信息，即使在 Address 表中没有关联的信息也要保留。此时可以用左外连接，将 Person 表放在 LEFT JOIN 的左边。
+
+```sql
+SELECT
+    FirstName,
+    LastName,
+    City,
+    State
+FROM
+    Person P
+    LEFT JOIN Address A
+    ON P.PersonId = A.PersonId;
+```
+
 ## SQL Schema
 
 ```sql
@@ -443,22 +464,6 @@ VALUES
     ( 1, 2, 'New York City', 'New York' );
 ```
 
-## Solution
-
-涉及到 Person 和 Address 两个表，在对这两个表执行连接操作时，因为要保留 Person 表中的信息，即使在 Address 表中没有关联的信息也要保留。此时可以用左外连接，将 Person 表放在 LEFT JOIN 的左边。
-
-```sql
-SELECT
-    FirstName,
-    LastName,
-    City,
-    State
-FROM
-    Person P
-    LEFT JOIN Address A
-    ON P.PersonId = A.PersonId;
-```
-
 # 181. Employees Earning More Than Their Managers
 
 https://leetcode.com/problems/employees-earning-more-than-their-managers/description/
@@ -480,6 +485,18 @@ Employee 表：
 
 查找薪资大于其经理薪资的员工信息。
 
+## Solution
+
+```sql
+SELECT
+    E1.NAME AS Employee
+FROM
+    Employee E1
+    INNER JOIN Employee E2
+    ON E1.ManagerId = E2.Id
+    AND E1.Salary > E2.Salary;
+```
+
 ## SQL Schema
 
 ```sql
@@ -495,18 +512,6 @@ VALUES
     ( 4, 'Max', 90000, NULL );
 ```
 
-## Solution
-
-```sql
-SELECT
-    E1.NAME AS Employee
-FROM
-    Employee E1
-    INNER JOIN Employee E2
-    ON E1.ManagerId = E2.Id
-    AND E1.Salary > E2.Salary;
-```
-
 # 183. Customers Who Never Order
 
 https://leetcode.com/problems/customers-who-never-order/description/
@@ -548,29 +553,6 @@ Orders 表：
 +-----------+
 ```
 
-## SQL Schema
-
-```sql
-DROP TABLE
-IF
-    EXISTS Customers;
-CREATE TABLE Customers ( Id INT, NAME VARCHAR ( 255 ) );
-DROP TABLE
-IF
-    EXISTS Orders;
-CREATE TABLE Orders ( Id INT, CustomerId INT );
-INSERT INTO Customers ( Id, NAME )
-VALUES
-    ( 1, 'Joe' ),
-    ( 2, 'Henry' ),
-    ( 3, 'Sam' ),
-    ( 4, 'Max' );
-INSERT INTO Orders ( Id, CustomerId )
-VALUES
-    ( 1, 3 ),
-    ( 2, 1 );
-```
-
 ## Solution
 
 左外链接
@@ -597,6 +579,29 @@ WHERE
     Id NOT IN ( SELECT CustomerId FROM Orders );
 ```
 
+## SQL Schema
+
+```sql
+DROP TABLE
+IF
+    EXISTS Customers;
+CREATE TABLE Customers ( Id INT, NAME VARCHAR ( 255 ) );
+DROP TABLE
+IF
+    EXISTS Orders;
+CREATE TABLE Orders ( Id INT, CustomerId INT );
+INSERT INTO Customers ( Id, NAME )
+VALUES
+    ( 1, 'Joe' ),
+    ( 2, 'Henry' ),
+    ( 3, 'Sam' ),
+    ( 4, 'Max' );
+INSERT INTO Orders ( Id, CustomerId )
+VALUES
+    ( 1, 3 ),
+    ( 2, 1 );
+```
+
 # 184. Department Highest Salary
 
 https://leetcode.com/problems/department-highest-salary/description/
@@ -638,25 +643,6 @@ Department 表：
 +------------+----------+--------+
 ```
 
-## SQL Schema
-
-```sql
-DROP TABLE IF EXISTS Employee;
-CREATE TABLE Employee ( Id INT, NAME VARCHAR ( 255 ), Salary INT, DepartmentId INT );
-DROP TABLE IF EXISTS Department;
-CREATE TABLE Department ( Id INT, NAME VARCHAR ( 255 ) );
-INSERT INTO Employee ( Id, NAME, Salary, DepartmentId )
-VALUES
-    ( 1, 'Joe', 70000, 1 ),
-    ( 2, 'Henry', 80000, 2 ),
-    ( 3, 'Sam', 60000, 2 ),
-    ( 4, 'Max', 90000, 1 );
-INSERT INTO Department ( Id, NAME )
-VALUES
-    ( 1, 'IT' ),
-    ( 2, 'Sales' );
-```
-
 ## Solution
 
 创建一个临时表，包含了部门员工的最大薪资。可以对部门进行分组，然后使用 MAX() 汇总函数取得最大薪资。
@@ -678,6 +664,26 @@ WHERE
     AND E.Salary = M.Salary;
 ```
 
+## SQL Schema
+
+```sql
+DROP TABLE IF EXISTS Employee;
+CREATE TABLE Employee ( Id INT, NAME VARCHAR ( 255 ), Salary INT, DepartmentId INT );
+DROP TABLE IF EXISTS Department;
+CREATE TABLE Department ( Id INT, NAME VARCHAR ( 255 ) );
+INSERT INTO Employee ( Id, NAME, Salary, DepartmentId )
+VALUES
+    ( 1, 'Joe', 70000, 1 ),
+    ( 2, 'Henry', 80000, 2 ),
+    ( 3, 'Sam', 60000, 2 ),
+    ( 4, 'Max', 90000, 1 );
+INSERT INTO Department ( Id, NAME )
+VALUES
+    ( 1, 'IT' ),
+    ( 2, 'Sales' );
+```
+
+
 # 176. Second Highest Salary
 
 https://leetcode.com/problems/second-highest-salary/description/
@@ -706,6 +712,15 @@ https://leetcode.com/problems/second-highest-salary/description/
 
 没有找到返回 null 而不是不返回数据。
 
+## Solution
+
+为了在没有查找到数据时返回 null，需要在查询结果外面再套一层 SELECT。
+
+```sql
+SELECT
+    ( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT 1, 1 ) SecondHighestSalary;
+```
+
 ## SQL Schema
 
 ```sql
@@ -720,25 +735,12 @@ VALUES
     ( 3, 300 );
 ```
 
-## Solution
-
-为了在没有查找到数据时返回 null，需要在查询结果外面再套一层 SELECT。
-
-```sql
-SELECT
-    ( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT 1, 1 ) SecondHighestSalary;
-```
-
 # 177. Nth Highest Salary
 
 ## Description
 
 查找工资第 N 高的员工。
 
-## SQL Schema
-
-同 176。
-
 ## Solution
 
 ```sql
@@ -750,6 +752,11 @@ RETURN ( SELECT ( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMI
 END
 ```
 
+## SQL Schema
+
+同 176。
+
+
 # 178. Rank Scores
 
 https://leetcode.com/problems/rank-scores/description/
@@ -786,23 +793,6 @@ https://leetcode.com/problems/rank-scores/description/
 +-------+------+
 ```
 
-## SQL Schema
-
-```sql
-DROP TABLE
-IF
-    EXISTS Scores;
-CREATE TABLE Scores ( Id INT, Score DECIMAL ( 3, 2 ) );
-INSERT INTO Scores ( Id, Score )
-VALUES
-    ( 1, 3.5 ),
-    ( 2, 3.65 ),
-    ( 3, 4.0 ),
-    ( 4, 3.85 ),
-    ( 5, 4.0 ),
-    ( 6, 3.65 );
-```
-
 ## Solution
 
 要统计某个 score 的排名，只要统计大于该 score 的 score 数量，然后加 1。
@@ -852,6 +842,23 @@ ORDER BY
     S1.score DESC;
 ```
 
+## SQL Schema
+
+```sql
+DROP TABLE
+IF
+    EXISTS Scores;
+CREATE TABLE Scores ( Id INT, Score DECIMAL ( 3, 2 ) );
+INSERT INTO Scores ( Id, Score )
+VALUES
+    ( 1, 3.5 ),
+    ( 2, 3.65 ),
+    ( 3, 4.0 ),
+    ( 4, 3.85 ),
+    ( 5, 4.0 ),
+    ( 6, 3.65 );
+```
+
 # 180. Consecutive Numbers
 
 https://leetcode.com/problems/consecutive-numbers/description/
@@ -884,6 +891,21 @@ https://leetcode.com/problems/consecutive-numbers/description/
 +-----------------+
 ```
 
+## Solution
+
+```sql
+SELECT
+    DISTINCT L1.num ConsecutiveNums
+FROM
+    Logs L1,
+    Logs L2,
+    Logs L3
+WHERE L1.id = l2.id - 1
+    AND L2.id = L3.id - 1
+    AND L1.num = L2.num
+    AND l2.num = l3.num;
+```
+
 ## SQL Schema
 
 ```sql
@@ -902,21 +924,6 @@ VALUES
     ( 7, 2 );
 ```
 
-## Solution
-
-```sql
-SELECT
-    DISTINCT L1.num ConsecutiveNums
-FROM
-    Logs L1,
-    Logs L2,
-    Logs L3
-WHERE L1.id = l2.id - 1
-    AND L2.id = L3.id - 1
-    AND L1.num = L2.num
-    AND l2.num = l3.num;
-```
-
 # 626. Exchange Seats
 
 https://leetcode.com/problems/exchange-seats/description/
@@ -951,22 +958,6 @@ seat 表存储着座位对应的学生。
 +---------+---------+
 ```
 
-## SQL Schema
-
-```sql
-DROP TABLE
-IF
-    EXISTS seat;
-CREATE TABLE seat ( id INT, student VARCHAR ( 255 ) );
-INSERT INTO seat ( id, student )
-VALUES
-    ( '1', 'Abbot' ),
-    ( '2', 'Doris' ),
-    ( '3', 'Emerson' ),
-    ( '4', 'Green' ),
-    ( '5', 'Jeames' );
-```
-
 ## Solution
 
 使用多个 union。
@@ -1004,6 +995,22 @@ ORDER BY
     id;
 ```
 
+## SQL Schema
+
+```sql
+DROP TABLE
+IF
+    EXISTS seat;
+CREATE TABLE seat ( id INT, student VARCHAR ( 255 ) );
+INSERT INTO seat ( id, student )
+VALUES
+    ( '1', 'Abbot' ),
+    ( '2', 'Doris' ),
+    ( '3', 'Emerson' ),
+    ( '4', 'Green' ),
+    ( '5', 'Jeames' );
+```
+