feat: add solutions to lc problem: No.1285 (#1798)

No.1285.Find the Start and End Number of Continuous Ranges

Author: yanglbme

.prettierrc

@@ -27,6 +27,7 @@
                 "solution/1100-1199/1174.Immediate Food Delivery II/Solution.sql",
                 "solution/1100-1199/1193.Monthly Transactions I/Solution.sql",
                 "solution/1200-1299/1205.Monthly Transactions II/Solution.sql",
+                "solution/1200-1299/1285.Find the Start and End Number of Continuous Ranges/Solution.sql",
                 "solution/1300-1399/1384.Total Sales Amount by Year/Solution.sql",
                 "solution/1300-1399/1322.Ads Performance/Solution.sql",
                 "solution/1300-1399/1393.Capital GainLoss/Solution.sql",

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

@@ -63,22 +63,51 @@ Logs 表：
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：分组 + 窗口函数**
+
+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。
+
+分组可以用以下两种方法实现：
+
+1. 通过计算每个日志与前一个日志的差值，如果差值为 $1$，则说明这两个日志是连续的，我们设置 $delta$ 为 $0$，否则设置为 $1$。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。
+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。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
-SELECT
-    MIN(log_id) AS start_id,
-    MAX(log_id) AS end_id
-FROM
-    (
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            log_id,
+            sum(delta) OVER (ORDER BY log_id) AS pid
+        FROM
+            (
+                SELECT
+                    log_id,
+                    if((log_id - lag(log_id) OVER (ORDER BY log_id)) = 1, 0, 1) AS delta
+                FROM Logs
+            ) AS t
+    )
+SELECT min(log_id) AS start_id, max(log_id) AS end_id
+FROM T
+GROUP BY pid;
+```
+
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
         SELECT
             log_id,
-            log_id - ROW_NUMBER() OVER (ORDER BY log_id) AS rk
+            log_id - row_number() OVER (ORDER BY log_id) AS pid
         FROM Logs
-    ) AS t
-GROUP BY rk;
+    )
+SELECT min(log_id) AS start_id, max(log_id) AS end_id
+FROM T
+GROUP BY pid;
 ```
 
 <!-- tabs:end -->

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

@@ -59,22 +59,51 @@ Number 10 is contained in the table.
 
 ## Solutions
 
+**Solution 1: Group By + Window Function**
+
+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.
+
+There are two ways to implement grouping:
+
+1. By calculating the difference between each log and the previous log, if the difference is $1$, then the two logs are continuous, and we set $delta$ to $0$, otherwise we set it to $1$. Then we take the prefix sum of $delta$ to obtain the grouping identifier for each row.
+2. By calculating the difference between the current log and its row number, we obtain the grouping identifier for each row.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
-SELECT
-    MIN(log_id) AS start_id,
-    MAX(log_id) AS end_id
-FROM
-    (
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            log_id,
+            sum(delta) OVER (ORDER BY log_id) AS pid
+        FROM
+            (
+                SELECT
+                    log_id,
+                    if((log_id - lag(log_id) OVER (ORDER BY log_id)) = 1, 0, 1) AS delta
+                FROM Logs
+            ) AS t
+    )
+SELECT min(log_id) AS start_id, max(log_id) AS end_id
+FROM T
+GROUP BY pid;
+```
+
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
         SELECT
             log_id,
-            log_id - ROW_NUMBER() OVER (ORDER BY log_id) AS rk
+            log_id - row_number() OVER (ORDER BY log_id) AS pid
         FROM Logs
-    ) AS t
-GROUP BY rk;
+    )
+SELECT min(log_id) AS start_id, max(log_id) AS end_id
+FROM T
+GROUP BY pid;
 ```
 
 <!-- tabs:end -->

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

@@ -1,11 +1,11 @@
-SELECT
-    MIN(log_id) AS start_id,
-    MAX(log_id) AS end_id
-FROM
-    (
+# Write your MySQL query statement below
+WITH
+    T AS (
         SELECT
             log_id,
-            log_id - ROW_NUMBER() OVER (ORDER BY log_id) AS rk
+            log_id-row_number() OVER (ORDER BY log_id) AS pid
         FROM Logs
-    ) AS t
-GROUP BY rk;
+    )
+SELECT min(log_id) AS start_id, max(log_id) AS end_id
+FROM T
+GROUP BY pid;