feat: add new sql solution

Author: yanglbme

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

@@ -57,31 +57,44 @@ Result 表：
 
 <!-- 这里可写通用的实现逻辑 -->
 
-**方法一：窗口函数**
+**方法一：两次连接**
 
-我们可以使用窗口函数 `LAG` 和 `LEAD` 来判断当前行的前一行和后一行是否与当前行的 `num` 相等，如果当前行与前一行的 `num` 相等，则记字段 $a$ 为 $1$，否则记为 $0$；如果当前行与后一行的 `num` 相等，则记字段 $b$ 为 $1$，否则记为 $0$。
+我们可以使用两次连接来解决这个问题。
 
-最后，我们只需要筛选出 $a = 1$ 且 $b = 1$ 的行，即为连续出现至少三次的数字。注意，我们需要使用 `DISTINCT` 关键字来对结果去重。
+我们首先进行一次自连接，连接条件是 `l1.num = l2.num` 并且 `l1.id = l2.id - 1`，这样我们就可以找出所有至少连续出现两次的数字。然后，我们再进行一次自连接，连接条件是 `l2.num = l3.num` 并且 `l2.id = l3.id - 1`，这样我们就可以找出所有至少连续出现三次的数字。最后，我们只需要筛选出去重的 `l2.num` 即可。
+
+**方法二：窗口函数**
+
+我们可以使用窗口函数 `LAG` 和 `LEAD` 来获取上一行的 `num` 和下一行的 `num`，记录在字段 $a$ 和 $b$ 中。最后，我们只需要筛选出 $a =num$ 并且 $b = num$ 的行，这些行就是至少连续出现三次的数字。注意，我们需要使用 `DISTINCT` 关键字来对结果去重。
 
 我们也可以对数字进行分组，具体做法是使用 `IF` 函数来判断当前行与前一行的 `num` 是否相等，如果相等则记为 $0$，否则记为 $1$，然后使用窗口函数 `SUM` 来计算前缀和，这样计算出的前缀和就是分组的标识。最后，我们只需要按照分组标识进行分组，然后筛选出每组中的行数大于等于 $3$ 的数字即可。同样，我们需要使用 `DISTINCT` 关键字来对结果去重。
 
 <!-- tabs:start -->
 
 ### **SQL**
 
+```sql
+# Write your MySQL query statement below
+SELECT DISTINCT l2.num AS ConsecutiveNums
+FROM
+    Logs AS l1
+    JOIN Logs AS l2 ON l1.id = l2.id - 1 AND l1.num = l2.num
+    JOIN Logs AS l3 ON l2.id = l3.id - 1 AND l2.num = l3.num;
+```
+
 ```sql
 # Write your MySQL query statement below
 WITH
     T AS (
         SELECT
             *,
-            num = (LAG(num) OVER (ORDER BY id)) AS a,
-            num = (LEAD(num) OVER (ORDER BY id)) AS b
+            LAG(num) OVER (ORDER BY id) AS a,
+            LEAD(num) OVER (ORDER BY id) AS b
         FROM Logs
     )
 SELECT DISTINCT num AS ConsecutiveNums
 FROM T
-WHERE a = 1 AND b = 1;
+WHERE a = num AND b = num;
 ```
 
 ```sql

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

@@ -53,31 +53,44 @@ Logs table:
 
 ## Solutions
 
-**Solution 1: Window Function**
+**Solution 1: Two Joins**
 
-We can use the window functions `LAG` and `LEAD` to determine whether the previous row and the next row of the current row are equal to the `num` of the current row. If the `num` of the current row is equal to the `num` of the previous row, we set the field $a$ to $1$, otherwise we set it to $0$; if the `num` of the current row is equal to the `num` of the next row, we set the field $b$ to $1$, otherwise we set it to $0$.
+We can use two joins to solve this problem.
 
-Finally, we only need to filter out the rows where $a = 1$ and $b = 1$, which are the numbers that appear at least three times in a row. Note that we need to use the `DISTINCT` keyword to remove duplicates from the results.
+First, we perform a self-join with the condition `l1.num = l2.num` and `l1.id = l2.id - 1`, so that we can find all numbers that appear at least twice in a row. Then, we perform another self-join with the condition `l2.num = l3.num` and `l2.id = l3.id - 1`, so that we can find all numbers that appear at least three times in a row. Finally, we only need to select the distinct `l2.num`.
+
+**Solution 2: Window Function**
+
+We can use the window functions `LAG` and `LEAD` to obtain the `num` of the previous row and the next row of the current row, and record them in the fields $a$ and $b$, respectively. Finally, we only need to filter out the rows where $a = num$ and $b = num$, which are the numbers that appear at least three times in a row. Note that we need to use the `DISTINCT` keyword to remove duplicates from the results.
 
 We can also group the numbers by using the `IF` function to determine whether the `num` of the current row is equal to the `num` of the previous row. If they are equal, we set it to $0$, otherwise we set it to $1$. Then, we use the window function `SUM` to calculate the prefix sum, which is the grouping identifier. Finally, we only need to group by the grouping identifier and filter out the numbers with a row count greater than or equal to $3$ in each group. Similarly, we need to use the `DISTINCT` keyword to remove duplicates from the results.
 
 <!-- tabs:start -->
 
 ### **SQL**
 
+```sql
+# Write your MySQL query statement below
+SELECT DISTINCT l2.num AS ConsecutiveNums
+FROM
+    Logs AS l1
+    JOIN Logs AS l2 ON l1.id = l2.id - 1 AND l1.num = l2.num
+    JOIN Logs AS l3 ON l2.id = l3.id - 1 AND l2.num = l3.num;
+```
+
 ```sql
 # Write your MySQL query statement below
 WITH
     T AS (
         SELECT
             *,
-            num = (LAG(num) OVER (ORDER BY id)) AS a,
-            num = (LEAD(num) OVER (ORDER BY id)) AS b
+            LAG(num) OVER (ORDER BY id) AS a,
+            LEAD(num) OVER (ORDER BY id) AS b
         FROM Logs
     )
 SELECT DISTINCT num AS ConsecutiveNums
 FROM T
-WHERE a = 1 AND b = 1;
+WHERE a = num AND b = num;
 ```
 
 ```sql