feat: update solutions to lc problems: No.1173,1445 (#1787)

Author: yanglbme

.prettierrc

@@ -31,6 +31,7 @@
                 "solution/1300-1399/1322.Ads Performance/Solution.sql",
                 "solution/1300-1399/1393.Capital GainLoss/Solution.sql",
                 "solution/1300-1399/1398.Customers Who Bought Products A and B but Not C/Solution.sql",
+                "solution/1400-1499/1445.Apples & Oranges/Solution.sql",
                 "solution/1400-1499/1479.Sales by Day of the Week/Solution.sql",
                 "solution/1500-1599/1501.Countries You Can Safely Invest In/Solution.sql",
                 "solution/1500-1599/1555.Bank Account Summary/Solution.sql",

solution/1100-1199/1173.Immediate Food Delivery I/README.md

@@ -58,14 +58,18 @@ Delivery 表:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：求和**
+
+我们可以用 `sum` 函数来统计即时订单的数量，然后除以总订单数即可。由于题目求的是百分比，所以需要乘以 100，最后我们用 `round` 函数保留两位小数。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
 SELECT
-    round(sum(order_date = customer_pref_delivery_date) * 100 / count(1), 2) AS immediate_percentage
+    round(sum(order_date = customer_pref_delivery_date) / count(1) * 100, 2) AS immediate_percentage
 FROM Delivery;
 ```
 

solution/1100-1199/1173.Immediate Food Delivery I/README_EN.md

@@ -54,14 +54,18 @@ Delivery table:
 
 ## Solutions
 
+**Solution 1: Sum**
+
+We can use the `sum` function to count the number of instant orders, and then divide it by the total number of orders. Since the problem requires a percentage, we need to multiply by 100. Finally, we can use the `round` function to keep two decimal places.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
 SELECT
-    round(sum(order_date = customer_pref_delivery_date) * 100 / count(1), 2) AS immediate_percentage
+    round(sum(order_date = customer_pref_delivery_date) / count(1) * 100, 2) AS immediate_percentage
 FROM Delivery;
 ```
 

solution/1100-1199/1173.Immediate Food Delivery I/Solution.sql

@@ -1,4 +1,4 @@
 # Write your MySQL query statement below
 SELECT
-    round(sum(order_date = customer_pref_delivery_date) * 100 / count(1), 2) AS immediate_percentage
+    round(sum(order_date = customer_pref_delivery_date) / count(1) * 100, 2) AS immediate_percentage
 FROM Delivery;

solution/1400-1499/1445.Apples & Oranges/README.md

@@ -67,7 +67,9 @@ Sales</code> 表:
 
 <!-- 这里可写通用的实现逻辑 -->
 
-`CASE WHEN` + `GROUP BY`。
+**方法一：分组求和**
+
+我们可以将数据按照日期分组，然后用 `sum` 函数求出每天苹果和桔子的销售差异。如果是苹果，我们就用正数表示，如果是桔子，我们就用负数表示。最后我们按照日期排序即可。
 
 <!-- tabs:start -->
 
@@ -76,16 +78,11 @@ Sales</code> 表:
 ```sql
 # Write your MySQL query statement below
 SELECT
-    sale_date AS SALE_DATE,
-    sum(
-        CASE
-            WHEN fruit = 'oranges' THEN -sold_num
-            ELSE sold_num
-        END
-    ) AS DIFF
+    sale_date,
+    sum(if(fruit = 'apples', sold_num, -sold_num)) AS diff
 FROM Sales
-GROUP BY sale_date
-ORDER BY sale_date;
+GROUP BY 1
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

solution/1400-1499/1445.Apples & Oranges/README_EN.md

@@ -62,23 +62,22 @@ Day 2020-05-04, 15 apples and 16 oranges were sold (Difference 15 - 16 = -1).
 
 ## Solutions
 
+**Solution 1: Group By + Sum**
+
+We can group the data by date, and then use the `sum` function to calculate the difference in sales between apples and oranges for each day. If it is an apple, we represent it with a positive number, and if it is an orange, we represent it with a negative number. Finally, we sort the data by date.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
 SELECT
-    sale_date AS SALE_DATE,
-    sum(
-        CASE
-            WHEN fruit = 'oranges' THEN -sold_num
-            ELSE sold_num
-        END
-    ) AS DIFF
+    sale_date,
+    sum(if(fruit = 'apples', sold_num, -sold_num)) AS diff
 FROM Sales
-GROUP BY sale_date
-ORDER BY sale_date;
+GROUP BY 1
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

solution/1400-1499/1445.Apples & Oranges/Solution.sql

@@ -1,12 +1,7 @@
 # Write your MySQL query statement below
 SELECT
-    sale_date AS SALE_DATE,
-    sum(
-        CASE
-            WHEN fruit = 'oranges' THEN -sold_num
-            ELSE sold_num
-        END
-    ) AS DIFF
+    sale_date,
+    sum(if(fruit = 'apples', sold_num, -sold_num)) AS diff
 FROM Sales
-GROUP BY sale_date
-ORDER BY sale_date;
+GROUP BY 1
+ORDER BY 1;