feat: add solution to lc problem: No.2474

No.2474.Customers With Strictly Increasing Purchases

Author: yanglbme

solution/2400-2499/2474.Customers With Strictly Increasing Purchases/README.md

@@ -90,7 +90,30 @@ Orders 表:
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+# Write your MySQL query statement below
+select
+    customer_id
+from
+    (
+        select
+            customer_id,
+            year(order_date),
+            sum(price) as total,
+            year(order_date) - rank() over(
+                partition by customer_id
+                order by
+                    sum(price)
+            ) as rk
+        from
+            Orders
+        group by
+            customer_id,
+            year(order_date)
+    ) t
+group by
+    customer_id
+having
+    count(distinct rk) = 1;
 ```
 
 <!-- tabs:end -->

solution/2400-2499/2474.Customers With Strictly Increasing Purchases/README_EN.md

@@ -85,7 +85,30 @@ Customer 3: The first year is 2017, and the last year is 2018
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+select
+    customer_id
+from
+    (
+        select
+            customer_id,
+            year(order_date),
+            sum(price) as total,
+            year(order_date) - rank() over(
+                partition by customer_id
+                order by
+                    sum(price)
+            ) as rk
+        from
+            Orders
+        group by
+            customer_id,
+            year(order_date)
+    ) t
+group by
+    customer_id
+having
+    count(distinct rk) = 1;
 ```
 
 <!-- tabs:end -->

solution/2400-2499/2474.Customers With Strictly Increasing Purchases/Solution.sql

@@ -0,0 +1,24 @@
+# Write your MySQL query statement below
+select
+    customer_id
+from
+    (
+        select
+            customer_id,
+            year(order_date),
+            sum(price) as total,
+            year(order_date) - rank() over(
+                partition by customer_id
+                order by
+                    sum(price)
+            ) as rk
+        from
+            Orders
+        group by
+            customer_id,
+            year(order_date)
+    ) t
+group by
+    customer_id
+having
+    count(distinct rk) = 1;