feat: add sql solutions to lc problems (#1173)

Author: yanglbme

.prettierignore

@@ -27,4 +27,6 @@ node_modules/
 /solution/1700-1799/1767.Find the Subtasks That Did Not Execute/Solution.sql
 /solution/2100-2199/2118.Build the Equation/Solution.sql
 /solution/2100-2199/2153.The Number of Passengers in Each Bus II/Solution.sql
-/solution/2200-2299/2205.The Number of Users That Are Eligible for Discount/Solution.sql
+/solution/2200-2299/2205.The Number of Users That Are Eligible for Discount/Solution.sql
+/solution/2200-2299/2252.Dynamic Pivoting of a Table/Solution.sql
+/solution/2200-2299/2253.Dynamic Unpivoting of a Table/Solution.sql

solution/0100-0199/0175.Combine Two Tables/README.md

@@ -80,20 +80,18 @@ addressId = 1 包含了 personId = 2 的地址信息。</pre>
 
 **方法一：左连接**
 
+我们可以使用左连接，将 `Person` 表左连接 `Address` 表，连接条件为 `Person.personId = Address.personId`，这样就可以得到每个人的姓、名、城市和州。如果 `personId` 的地址不在 `Address` 表中，则报告为空 `null`。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    firstName,
-    lastName,
-    city,
-    state
+SELECT firstName, lastName, city, state
 FROM
-    Person AS p
-    LEFT JOIN Address AS a ON p.personId = a.personId;
+    Person
+    LEFT JOIN Address USING (personId);
 ```
 
 <!-- tabs:end -->

solution/0100-0199/0175.Combine Two Tables/README_EN.md

@@ -82,14 +82,10 @@ addressId = 1 contains information about the address of personId = 2.
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    firstName,
-    lastName,
-    city,
-    state
+SELECT firstName, lastName, city, state
 FROM
-    Person AS p
-    LEFT JOIN Address AS a ON p.personId = a.personId;
+    Person
+    LEFT JOIN Address USING (personId);
 ```
 
 <!-- tabs:end -->

solution/0100-0199/0175.Combine Two Tables/Solution.sql

@@ -1,9 +1,5 @@
 # Write your MySQL query statement below
-SELECT
-    firstName,
-    lastName,
-    city,
-    state
+SELECT firstName, lastName, city, state
 FROM
-    Person AS p
-    LEFT JOIN Address AS a ON p.personId = a.personId;
+    Person
+    LEFT JOIN Address USING (personId);

solution/0100-0199/0176.Second Highest Salary/README.md

@@ -73,12 +73,22 @@ Employee 表：
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：使用 LIMIT 语句和子查询**
+
+我们可以按照薪水降序排列，然后使用 `LIMIT` 语句来获取第二高的薪水，如果不存在第二高的薪水，那么就返回 `null`。
+
+**方法二：使用 MAX() 函数和子查询**
+
+使用 `MAX()` 函数，从小于 `MAX()` 的 Salary 中挑选最大值 `MAX()` 即可。
+
+**方法三：使用 IFNULL() 和窗口函数**
+
+我们也可以先通过 `dense_rank()` 函数计算出每个员工的薪水排名，然后再筛选出排名为 $2$ 的员工即可，如果不存在第二高的薪水，那么就返回 `null`。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
-解法 1：使用 LIMIT 语句和子查询。
-
 ```sql
 # Write your MySQL query statement below
 SELECT
@@ -90,8 +100,6 @@ SELECT
     ) AS SecondHighestSalary;
 ```
 
-解法 2：使用 `MAX()` 函数，从小于 `MAX()` 的 Salary 中挑选最大值 `MAX()` 即可。
-
 ```sql
 # Write your MySQL query statement below
 SELECT MAX(Salary) AS SecondHighestSalary
@@ -103,4 +111,20 @@ WHERE
     );
 ```
 
+```sql
+# Write your MySQL query statement below
+WITH
+    S AS (
+        SELECT salary, dense_rank() OVER (ORDER BY salary DESC) AS rk
+        FROM Employee
+    )
+SELECT
+    ifnull(
+        SELECT salary
+        FROM S
+        WHERE rk = 2
+        LIMIT 1, NULL
+    ) AS SecondHighestSalary;
+```
+
 <!-- tabs:end -->

solution/0100-0199/0176.Second Highest Salary/README_EN.md

@@ -64,12 +64,16 @@ Employee table:
 
 ## Solutions
 
+**Solution 1: Use Sub Query and LIMIT**
+
+**Solution 2: Use `MAX()` function**
+
+**Solution 3: Use `IFNULL()` and window function**
+
 <!-- tabs:start -->
 
 ### **SQL**
 
-Solution 1: Use Sub Query and LIMIT.
-
 ```sql
 # Write your MySQL query statement below
 SELECT
@@ -81,8 +85,6 @@ SELECT
     ) AS SecondHighestSalary;
 ```
 
-Solution 2: Use `MAX()` function.
-
 ```sql
 # Write your MySQL query statement below
 SELECT MAX(Salary) AS SecondHighestSalary
@@ -94,4 +96,20 @@ WHERE
     );
 ```
 
+```sql
+# Write your MySQL query statement below
+WITH
+    S AS (
+        SELECT salary, dense_rank() OVER (ORDER BY salary DESC) AS rk
+        FROM Employee
+    )
+SELECT
+    ifnull(
+        SELECT salary
+        FROM S
+        WHERE rk = 2
+        LIMIT 1, NULL
+    ) AS SecondHighestSalary;
+```
+
 <!-- tabs:end -->

solution/0100-0199/0180.Consecutive Numbers/README.md

@@ -60,26 +60,23 @@ Result 表：
 
 ### **SQL**
 
-```sql
-SELECT DISTINCT (Num) AS ConsecutiveNums
-FROM Logs AS Curr
-WHERE
-    Num = (SELECT Num FROM Logs WHERE id = Curr.id - 1)
-    AND Num = (SELECT Num FROM Logs WHERE id = Curr.id - 2);
-```
-
 ```sql
 # Write your MySQL query statement below
-SELECT DISTINCT l1.num AS ConsecutiveNums
-FROM
-    logs AS l1,
-    logs AS l2,
-    logs AS l3
-WHERE
-    l1.id = l2.id - 1
-    AND l2.id = l3.id - 1
-    AND l1.num = l2.num
-    AND l2.num = l3.num;
+WITH
+    t AS (
+        SELECT
+            *,
+            CASE
+                WHEN (lag(num) OVER (ORDER BY id)) = num THEN 0
+                ELSE 1
+            END AS mark
+        FROM Logs
+    ),
+    p AS (SELECT num, sum(mark) OVER (ORDER BY id) AS gid FROM t)
+SELECT DISTINCT num AS ConsecutiveNums
+FROM p
+GROUP BY gid
+HAVING count(1) >= 3;
 ```
 
 <!-- tabs:end -->

solution/0100-0199/0180.Consecutive Numbers/README_EN.md

@@ -57,26 +57,23 @@ Logs table:
 
 ### **SQL**
 
-```sql
-SELECT DISTINCT (Num) AS ConsecutiveNums
-FROM Logs AS Curr
-WHERE
-    Num = (SELECT Num FROM Logs WHERE id = Curr.id - 1)
-    AND Num = (SELECT Num FROM Logs WHERE id = Curr.id - 2);
-```
-
 ```sql
 # Write your MySQL query statement below
-SELECT DISTINCT l1.num AS ConsecutiveNums
-FROM
-    logs AS l1,
-    logs AS l2,
-    logs AS l3
-WHERE
-    l1.id = l2.id - 1
-    AND l2.id = l3.id - 1
-    AND l1.num = l2.num
-    AND l2.num = l3.num;
+WITH
+    t AS (
+        SELECT
+            *,
+            CASE
+                WHEN (lag(num) OVER (ORDER BY id)) = num THEN 0
+                ELSE 1
+            END AS mark
+        FROM Logs
+    ),
+    p AS (SELECT num, sum(mark) OVER (ORDER BY id) AS gid FROM t)
+SELECT DISTINCT num AS ConsecutiveNums
+FROM p
+GROUP BY gid
+HAVING count(1) >= 3;
 ```
 
 <!-- tabs:end -->

solution/0100-0199/0180.Consecutive Numbers/Solution.sql

@@ -1,5 +1,16 @@
-SELECT DISTINCT (Num) AS ConsecutiveNums
-FROM Logs AS Curr
-WHERE
-    Num = (SELECT Num FROM Logs WHERE id = Curr.id - 1)
-    AND Num = (SELECT Num FROM Logs WHERE id = Curr.id - 2);
+# Write your MySQL query statement below
+WITH
+    t AS (
+        SELECT
+            *,
+            CASE
+                WHEN (lag(num) OVER (ORDER BY id)) = num THEN 0
+                ELSE 1
+            END AS mark
+        FROM Logs
+    ),
+    p AS (SELECT num, sum(mark) OVER (ORDER BY id) AS gid FROM t)
+SELECT DISTINCT num AS ConsecutiveNums
+FROM p
+GROUP BY gid
+HAVING count(1) >= 3;

solution/2200-2299/2252.Dynamic Pivoting of a Table/README.md

@@ -76,7 +76,23 @@ Products 表:
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+CREATE PROCEDURE PivotProducts()
+BEGIN
+	# Write your MySQL query statement below.
+	SET group_concat_max_len = 5000;
+    SELECT GROUP_CONCAT(DISTINCT 'MAX(CASE WHEN store = \'',
+               store,
+               '\' THEN price ELSE NULL END) AS ',
+               store
+               ORDER BY store) INTO @sql
+    FROM Products;
+    SET @sql =  CONCAT('SELECT product_id, ',
+                    @sql,
+                    ' FROM Products GROUP BY product_id');
+    PREPARE stmt FROM @sql;
+    EXECUTE stmt;
+    DEALLOCATE PREPARE stmt;
+END
 ```
 
 <!-- tabs:end -->