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Commit 0c7a735

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feat: update sql solutions to lc problems (#1188)
* No.0585.Investments in 2016 * No.0586.Customer Placing the Largest Number of Orders * No.0595.Big Countries * No.0596.Classes More Than 5 Students * No.0597.Friend Requests I Overall Acceptance Rate
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‎solution/0000-0099/0016.3Sum Closest/README.md‎

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**方法一:排序 + 双指针**
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将数组排序,然后遍历数组,对于每个元素 $nums[i],ドル我们使用指针 $j$ 和 $k$ 分别指向 $i+1$ 和 $n-1,ドル计算三数之和,如果三数之和等于 $target,ドル则直接返回 $target,ドル否则根据与 $target$ 的差值更新答案。如果三数之和大于 $target,ドル则将 $k$ 向左移动一位,否则将 $j$ 向右移动一位。
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我们将数组排序,然后遍历数组,对于每个元素 $nums[i],ドル我们使用指针 $j$ 和 $k$ 分别指向 $i+1$ 和 $n-1,ドル计算三数之和,如果三数之和等于 $target,ドル则直接返回 $target,ドル否则根据与 $target$ 的差值更新答案。如果三数之和大于 $target,ドル则将 $k$ 向左移动一位,否则将 $j$ 向右移动一位。
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时间复杂度 $O(n^2),ドル空间复杂度 $O(\log n)$。其中 $n$ 为数组长度。
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‎solution/0500-0599/0585.Investments in 2016/README.md‎

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

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

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

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

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LIMIT 1;
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```
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SQL Server
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```sql
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/* Write your T-SQL query statement below */
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SELECT TOP 1
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customer_number
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FROM

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

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```sql
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# Write your MySQL query statement below
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SELECT
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customer_number
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FROM orders
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GROUP BY customer_number
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SELECT customer_number
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FROM Orders
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GROUP BY 1
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ORDER BY count(1) DESC
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LIMIT 1;
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```
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SQL Server
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```sql
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/* Write your T-SQL query statement below */
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SELECT TOP 1
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customer_number
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FROM
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# Write your MySQL query statement below
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SELECT
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customer_number
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FROM orders
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GROUP BY customer_number
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SELECT customer_number
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FROM Orders
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GROUP BY 1
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ORDER BY count(1) DESC
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LIMIT 1;

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

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:使用 WHERE + OR**
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我们可以使用 `WHERE` + `OR` 查询出所有符合条件的国家。
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**方法二:使用 UNION**
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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。
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<!-- tabs:start -->
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### **SQL**
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```sql
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SELECT
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name,
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population,
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area
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FROM world
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WHERE area > 3000000 OR population > 25000000;
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# Write your MySQL query statement below
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SELECT name, population, area
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FROM World
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WHERE area >= 3000000 OR population >= 25000000;
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```
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```sql
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# Write your MySQL query statement below
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SELECT name, population, area
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FROM World
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WHERE area >= 3000000
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UNION
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SELECT name, population, area
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FROM World
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WHERE population >= 25000000;
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```
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<!-- tabs:end -->

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

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

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