feat: update sql solutions to lc problems

solution/0000-0099/0016.3Sum Closest/README.md

@@ -45,7 +45,7 @@
 
 **方法一：排序 + 双指针**
 
-将数组排序，然后遍历数组，对于每个元素 $nums[i]$，我们使用指针 $j$ 和 $k$ 分别指向 $i+1$ 和 $n-1$，计算三数之和，如果三数之和等于 $target$，则直接返回 $target$，否则根据与 $target$ 的差值更新答案。如果三数之和大于 $target$，则将 $k$ 向左移动一位，否则将 $j$ 向右移动一位。
+我们将数组排序，然后遍历数组，对于每个元素 $nums[i]$，我们使用指针 $j$ 和 $k$ 分别指向 $i+1$ 和 $n-1$，计算三数之和，如果三数之和等于 $target$，则直接返回 $target$，否则根据与 $target$ 的差值更新答案。如果三数之和大于 $target$，则将 $k$ 向左移动一位，否则将 $j$ 向右移动一位。
 
 时间复杂度 $O(n^2)$，空间复杂度 $O(\log n)$。其中 $n$ 为数组长度。
 

solution/0500-0599/0585.Investments in 2016/README.md

@@ -80,17 +80,16 @@ tiv_2015 值为 10 与第三条和第四条记录相同，且其位置是唯一
 ```sql
 # Write your MySQL query statement below
 WITH
-    t AS (
+    T AS (
         SELECT
             tiv_2016,
             count(pid) OVER (PARTITION BY tiv_2015) AS cnt1,
-            count(pid) OVER (PARTITION BY concat(lat, lon)) AS cnt2
+            count(pid) OVER (PARTITION BY concat(lat, '-', lon)) AS cnt2
         FROM Insurance
     )
-SELECT
-    round(sum(TIV_2016), 2) AS tiv_2016
-FROM t
-WHERE cnt1 != 1 AND cnt2 = 1;
+SELECT round(ifnull(sum(tiv_2016), 0), 2) AS tiv_2016
+FROM T
+WHERE cnt1 > 1 AND cnt2 = 1;
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0585.Investments in 2016/README_EN.md

@@ -74,17 +74,16 @@ So, the result is the sum of tiv_2016 of the first and last record, which is 45.
 ```sql
 # Write your MySQL query statement below
 WITH
-    t AS (
+    T AS (
         SELECT
             tiv_2016,
             count(pid) OVER (PARTITION BY tiv_2015) AS cnt1,
-            count(pid) OVER (PARTITION BY concat(lat, lon)) AS cnt2
+            count(pid) OVER (PARTITION BY concat(lat, '-', lon)) AS cnt2
         FROM Insurance
     )
-SELECT
-    round(sum(TIV_2016), 2) AS tiv_2016
-FROM t
-WHERE cnt1 != 1 AND cnt2 = 1;
+SELECT round(ifnull(sum(tiv_2016), 0), 2) AS tiv_2016
+FROM T
+WHERE cnt1 > 1 AND cnt2 = 1;
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0585.Investments in 2016/Solution.sql

@@ -1,13 +1,12 @@
 # Write your MySQL query statement below
 WITH
-    t AS (
+    T AS (
         SELECT
             tiv_2016,
             count(pid) OVER (PARTITION BY tiv_2015) AS cnt1,
-            count(pid) OVER (PARTITION BY concat(lat, lon)) AS cnt2
+            count(pid) OVER (PARTITION BY concat(lat, '-', lon)) AS cnt2
         FROM Insurance
     )
-SELECT
-    round(sum(TIV_2016), 2) AS tiv_2016
-FROM t
-WHERE cnt1 != 1 AND cnt2 = 1;
+SELECT round(ifnull(sum(tiv_2016), 0), 2) AS tiv_2016
+FROM T
+WHERE cnt1 > 1 AND cnt2 = 1;

solution/0500-0599/0586.Customer Placing the Largest Number of Orders/README.md

@@ -75,9 +75,8 @@ ORDER BY count(1) DESC
 LIMIT 1;
 ```
 
-SQL Server
-
 ```sql
+/* Write your T-SQL query statement below */
 SELECT TOP 1
     customer_number
 FROM

solution/0500-0599/0586.Customer Placing the Largest Number of Orders/README_EN.md

@@ -61,17 +61,15 @@ So the result is customer_number 3.
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    customer_number
-FROM orders
-GROUP BY customer_number
+SELECT customer_number
+FROM Orders
+GROUP BY 1
 ORDER BY count(1) DESC
 LIMIT 1;
 ```
 
-SQL Server
-
 ```sql
+/* Write your T-SQL query statement below */
 SELECT TOP 1
     customer_number
 FROM

solution/0500-0599/0586.Customer Placing the Largest Number of Orders/Solution.sql

@@ -1,7 +1,6 @@
 # Write your MySQL query statement below
-SELECT
-    customer_number
-FROM orders
-GROUP BY customer_number
+SELECT customer_number
+FROM Orders
+GROUP BY 1
 ORDER BY count(1) DESC
 LIMIT 1;

solution/0500-0599/0595.Big Countries/README.md

@@ -72,17 +72,34 @@ World 表：
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：使用 WHERE + OR**
+
+我们可以使用 `WHERE` + `OR` 查询出所有符合条件的国家。
+
+**方法二：使用 UNION**
+
+我们可以查询出所有面积大于等于 300 万平方公里的国家，然后再查询出所有人口大于等于 2500 万的国家，最后使用 `UNION` 将两个结果集合并起来。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
-SELECT
-    name,
-    population,
-    area
-FROM world
-WHERE area > 3000000 OR population > 25000000;
+# Write your MySQL query statement below
+SELECT name, population, area
+FROM World
+WHERE area >= 3000000 OR population >= 25000000;
+```
+
+```sql
+# Write your MySQL query statement below
+SELECT name, population, area
+FROM World
+WHERE area >= 3000000
+UNION
+SELECT name, population, area
+FROM World
+WHERE population >= 25000000;
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0595.Big Countries/README_EN.md

@@ -66,12 +66,21 @@ World table:
 ### **SQL**
 
 ```sql
-SELECT
-    name,
-    population,
-    area
-FROM world
-WHERE area > 3000000 OR population > 25000000;
+# Write your MySQL query statement below
+SELECT name, population, area
+FROM World
+WHERE area >= 3000000 OR population >= 25000000;
+```
+
+```sql
+# Write your MySQL query statement below
+SELECT name, population, area
+FROM World
+WHERE area >= 3000000
+UNION
+SELECT name, population, area
+FROM World
+WHERE population >= 25000000;
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0595.Big Countries/Solution.sql

@@ -1,6 +1,4 @@
-SELECT
-    name,
-    population,
-    area
-FROM world
-WHERE area > 3000000 OR population > 25000000;
+# Write your MySQL query statement below
+SELECT name, population, area
+FROM World
+WHERE area >= 3000000 OR population >= 25000000;

solution/0500-0599/0596.Classes More Than 5 Students/README.md

@@ -68,11 +68,11 @@ Courses table:
 ### **SQL**
 
 ```sql
-SELECT
-    class
-FROM courses
+# Write your MySQL query statement below
+SELECT class
+FROM Courses
 GROUP BY class
-HAVING COUNT(class) >= 5;
+HAVING count(1) >= 5;
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0596.Classes More Than 5 Students/README_EN.md

@@ -64,11 +64,11 @@ Courses table:
 ### **SQL**
 
 ```sql
-SELECT
-    class
-FROM courses
+# Write your MySQL query statement below
+SELECT class
+FROM Courses
 GROUP BY class
-HAVING COUNT(class) >= 5;
+HAVING count(1) >= 5;
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0596.Classes More Than 5 Students/Solution.sql

@@ -1,5 +1,5 @@
-SELECT
-    class
-FROM courses
+# Write your MySQL query statement below
+SELECT class
+FROM Courses
 GROUP BY class
-HAVING COUNT(class) >= 5;
+HAVING count(1) >= 5;

solution/0500-0599/0597.Friend Requests I Overall Acceptance Rate/README.md

@@ -102,19 +102,19 @@ RequestAccepted 表：
 ### **SQL**
 
 ```sql
+# Write your MySQL query statement below
 SELECT
-    IFNULL(
-        ROUND(
+    round(
+        ifnull(
             (
-                SELECT COUNT(DISTINCT requester_id, accepter_id)
+                SELECT count(DISTINCT requester_id, accepter_id)
                 FROM RequestAccepted
             ) / (
-                SELECT COUNT(DISTINCT sender_id, send_to_id)
-                FROM FriendRequest
+                SELECT count(DISTINCT sender_id, send_to_id) FROM FriendRequest
             ),
-            2
+            0
         ),
-        0.00
+        2
     ) AS accept_rate;
 ```
 

solution/0500-0599/0597.Friend Requests I Overall Acceptance Rate/README_EN.md

@@ -98,19 +98,19 @@ There are 4 unique accepted requests, and there are 5 requests in total. So the
 ### **SQL**
 
 ```sql
+# Write your MySQL query statement below
 SELECT
-    IFNULL(
-        ROUND(
+    round(
+        ifnull(
             (
-                SELECT COUNT(DISTINCT requester_id, accepter_id)
+                SELECT count(DISTINCT requester_id, accepter_id)
                 FROM RequestAccepted
             ) / (
-                SELECT COUNT(DISTINCT sender_id, send_to_id)
-                FROM FriendRequest
+                SELECT count(DISTINCT sender_id, send_to_id) FROM FriendRequest
             ),
-            2
+            0
         ),
-        0.00
+        2
     ) AS accept_rate;
 ```
 

solution/0500-0599/0597.Friend Requests I Overall Acceptance Rate/Solution.sql

@@ -1,14 +1,14 @@
+# Write your MySQL query statement below
 SELECT
-    IFNULL(
-        ROUND(
+    round(
+        ifnull(
             (
-                SELECT COUNT(DISTINCT requester_id, accepter_id)
+                SELECT count(DISTINCT requester_id, accepter_id)
                 FROM RequestAccepted
             ) / (
-                SELECT COUNT(DISTINCT sender_id, send_to_id)
-                FROM FriendRequest
+                SELECT count(DISTINCT sender_id, send_to_id) FROM FriendRequest
             ),
-            2
+            0
         ),
-        0.00
+        2
     ) AS accept_rate;