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Commit 242c377

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feat: add solutions to lc problem: No.1285 (doocs#1798)
No.1285.Find the Start and End Number of Continuous Ranges
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‎.prettierrc‎

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"solution/1100-1199/1174.Immediate Food Delivery II/Solution.sql",
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"solution/1100-1199/1193.Monthly Transactions I/Solution.sql",
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"solution/1200-1299/1205.Monthly Transactions II/Solution.sql",
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"solution/1200-1299/1285.Find the Start and End Number of Continuous Ranges/Solution.sql",
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"solution/1300-1399/1384.Total Sales Amount by Year/Solution.sql",
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"solution/1300-1399/1322.Ads Performance/Solution.sql",
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"solution/1300-1399/1393.Capital GainLoss/Solution.sql",

‎solution/1200-1299/1285.Find the Start and End Number of Continuous Ranges/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:分组 + 窗口函数**
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我们需要想办法将一段连续的日志分到同一组,然后对每一组进行聚合操作,得到每一组的开始日志和结束日志。
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分组可以用以下两种方法实现:
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1. 通过计算每个日志与前一个日志的差值,如果差值为 1ドル,ドル则说明这两个日志是连续的,我们设置 $delta$ 为 0ドル,ドル否则设置为 1ドル$。然后我们对 $delta$ 求前缀和,得到的结果就是每一行的分组的标识符。
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2. 通过计算当前行的日志减去当前行的行号,得到的结果就是每一行的分组的标识符。
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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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MIN(log_id) AS start_id,
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MAX(log_id) AS end_id
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FROM
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(
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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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SELECT
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log_id,
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sum(delta) OVER (ORDER BY log_id) AS pid
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FROM
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(
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SELECT
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log_id,
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if((log_id - lag(log_id) OVER (ORDER BY log_id)) = 1, 0, 1) AS delta
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FROM Logs
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) AS t
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)
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SELECT min(log_id) AS start_id, max(log_id) AS end_id
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FROM T
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GROUP BY pid;
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```
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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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SELECT
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log_id,
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log_id - ROW_NUMBER() OVER (ORDER BY log_id) AS rk
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log_id - row_number() OVER (ORDER BY log_id) AS pid
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FROM Logs
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) AS t
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GROUP BY rk;
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)
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SELECT min(log_id) AS start_id, max(log_id) AS end_id
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FROM T
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GROUP BY pid;
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```
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<!-- tabs:end -->

‎solution/1200-1299/1285.Find the Start and End Number of Continuous Ranges/README_EN.md‎

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## Solutions
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**Solution 1: Group By + Window Function**
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We need to find a way to group a continuous sequence of logs into the same group, and then aggregate each group to obtain the start and end logs of each group.
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There are two ways to implement grouping:
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1. By calculating the difference between each log and the previous log, if the difference is 1ドル,ドル then the two logs are continuous, and we set $delta$ to 0ドル,ドル otherwise we set it to 1ドル$. Then we take the prefix sum of $delta$ to obtain the grouping identifier for each row.
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2. By calculating the difference between the current log and its row number, we obtain the grouping identifier for each row.
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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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MIN(log_id) AS start_id,
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MAX(log_id) AS end_id
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FROM
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(
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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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SELECT
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log_id,
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sum(delta) OVER (ORDER BY log_id) AS pid
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FROM
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(
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SELECT
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log_id,
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if((log_id - lag(log_id) OVER (ORDER BY log_id)) = 1, 0, 1) AS delta
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FROM Logs
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) AS t
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)
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SELECT min(log_id) AS start_id, max(log_id) AS end_id
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FROM T
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GROUP BY pid;
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```
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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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SELECT
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log_id,
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log_id - ROW_NUMBER() OVER (ORDER BY log_id) AS rk
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log_id - row_number() OVER (ORDER BY log_id) AS pid
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FROM Logs
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) AS t
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GROUP BY rk;
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)
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SELECT min(log_id) AS start_id, max(log_id) AS end_id
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FROM T
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GROUP BY pid;
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```
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<!-- tabs:end -->
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SELECT
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MIN(log_id) AS start_id,
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MAX(log_id) AS end_id
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FROM
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(
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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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SELECT
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log_id,
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log_id- ROW_NUMBER() OVER (ORDER BY log_id) AS rk
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log_id-row_number() OVER (ORDER BY log_id) AS pid
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FROM Logs
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) AS t
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GROUP BY rk;
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)
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SELECT min(log_id) AS start_id, max(log_id) AS end_id
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FROM T
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GROUP BY pid;

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