feat: update solutions to lc problems: No.0197,0584,1581,1683,1757

solution/0100-0199/0197.Rising Temperature/README.md

@@ -61,6 +61,10 @@ Weather 表：</code>
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：自连接 + DATEDIFF/SUBDATE 函数**
+
+我们可以通过自连接的方式，将 `Weather` 表中的每一行与它的前一行进行比较，如果温度更高，并且日期相差一天，那么就是我们要找的结果。
+
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 ### **SQL**
@@ -75,13 +79,12 @@ FROM
 ```
 
 ```sql
-SELECT
-	w2.id AS Id
+# Write your MySQL query statement below
+SELECT w1.id
 FROM
-	weather AS w1
-	JOIN weather AS w2 ON DATE_ADD( w1.recordDate, INTERVAL 1 DAY) = w2.recordDate
-WHERE
-	w1.temperature < w2.temperature
+    Weather AS w1
+    JOIN Weather AS w2
+        ON SUBDATE(w1.recordDate, 1) = w2.recordDate AND w1.temperature > w2.temperature;
 ```
 
 <!-- tabs:end -->

solution/0100-0199/0197.Rising Temperature/README_EN.md

@@ -54,6 +54,10 @@ In 2015-01-04, the temperature was higher than the previous day (20 -> 30).
 
 ## Solutions
 
+**Solution 1: Self-Join + DATEDIFF/SUBDATE Function**
+
+We can use self-join to compare each row in the `Weather` table with its previous row. If the temperature is higher and the date difference is one day, then it is the result we are looking for.
+
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 ### **SQL**
@@ -68,13 +72,12 @@ FROM
 ```
 
 ```sql
-SELECT
-	w2.id AS Id
+# Write your MySQL query statement below
+SELECT w1.id
 FROM
-	weather AS w1
-	JOIN weather AS w2 ON DATE_ADD( w1.recordDate, INTERVAL 1 DAY) = w2.recordDate
-WHERE
-	w1.temperature < w2.temperature
+    Weather AS w1
+    JOIN Weather AS w2
+        ON SUBDATE(w1.recordDate, 1) = w2.recordDate AND w1.temperature > w2.temperature;
 ```
 
 <!-- tabs:end -->

solution/0100-0199/0197.Rising Temperature/Solution.sql

@@ -3,4 +3,4 @@ SELECT w1.id
 FROM
     Weather AS w1
     JOIN Weather AS w2
-        ON DATEDIFF(w1.recordDate, w2.recordDate) = 1 AND w1.temperature > w2.temperature;
+        ON SUBDATE(w1.recordDate, 1) = w2.recordDate AND w1.temperature > w2.temperature;

solution/0500-0599/0584.Find Customer Referee/README.md

@@ -56,6 +56,10 @@ Customer 表:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：条件过滤**
+
+我们可以直接筛选出 `referee_id` 不为 `2` 的客户姓名。注意，`referee_id` 为 `NULL` 的客户也应该被筛选出来。
+
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 ### **SQL**

solution/0500-0599/0584.Find Customer Referee/README_EN.md

@@ -55,6 +55,10 @@ Customer table:
 
 ## Solutions
 
+**Solution 1: Conditional Filtering**
+
+We can directly filter out the customer names whose `referee_id` is not `2`. Note that the customers whose `referee_id` is `NULL` should also be filtered out.
+
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 ### **SQL**

solution/1500-1599/1581.Customer Who Visited but Did Not Make Any Transactions/README.md

@@ -91,33 +91,36 @@ ID = 96 的顾客曾经去过购物中心，并且没有进行任何交易。
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：子查询 + 分组统计**
+
+我们可以使用子查询，先找出所有没有进行交易的 `visit_id`，然后按照 `customer_id` 进行分组，统计每个顾客的没有进行交易的次数。
+
+**方法二：左连接 + 分组统计**
+
+我们也可以使用左连接，将 `Visits` 表和 `Transactions` 表按照 `visit_id` 进行连接，然后筛选出 `amount` 为 `NULL` 的记录，按照 `customer_id` 进行分组，统计每个顾客的没有进行交易的次数。
+
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 ### **SQL**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-SELECT
-    customer_id,
-    COUNT(*) AS count_no_trans
+# Write your MySQL query statement below
+SELECT customer_id, COUNT(1) AS count_no_trans
 FROM Visits
-WHERE
-    visit_id NOT IN (
-        SELECT visit_id
-        FROM Transactions
-    )
-GROUP BY customer_id;
+WHERE visit_id NOT IN (SELECT visit_id FROM Transactions)
+GROUP BY 1;
 ```
 
 ```sql
 # Write your MySQL query statement below
 SELECT customer_id, COUNT(1) AS count_no_trans
 FROM
-    Visits AS v
-    LEFT JOIN Transactions AS t ON v.visit_id = t.visit_id
+    Visits
+    LEFT JOIN Transactions USING (visit_id)
 WHERE amount IS NULL
-GROUP BY customer_id;
+GROUP BY 1;
 ```
 
 <!-- tabs:end -->

solution/1500-1599/1581.Customer Who Visited but Did Not Make Any Transactions/README_EN.md

@@ -87,33 +87,36 @@ As we can see, users with IDs 30 and 96 visited the mall one time without making
 
 ## Solutions
 
+**Solution 1: Subquery + Grouping**
+
+We can use a subquery to first find all `visit_id`s that have not made any transactions, and then group by `customer_id` to count the number of times each customer has not made any transactions.
+
+**Solution 2: Left Join + Grouping**
+
+We can also use a left join to join the `Visits` table and the `Transactions` table on `visit_id`, and then filter out the records where `amount` is `NULL`. After that, we can group by `customer_id` to count the number of times each customer has not made any transactions.
+
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 ### **SQL**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-SELECT
-    customer_id,
-    COUNT(*) AS count_no_trans
+# Write your MySQL query statement below
+SELECT customer_id, COUNT(1) AS count_no_trans
 FROM Visits
-WHERE
-    visit_id NOT IN (
-        SELECT visit_id
-        FROM Transactions
-    )
-GROUP BY customer_id;
+WHERE visit_id NOT IN (SELECT visit_id FROM Transactions)
+GROUP BY 1;
 ```
 
 ```sql
 # Write your MySQL query statement below
 SELECT customer_id, COUNT(1) AS count_no_trans
 FROM
-    Visits AS v
-    LEFT JOIN Transactions AS t ON v.visit_id = t.visit_id
+    Visits
+    LEFT JOIN Transactions USING (visit_id)
 WHERE amount IS NULL
-GROUP BY customer_id;
+GROUP BY 1;
 ```
 
 <!-- tabs:end -->

solution/1500-1599/1581.Customer Who Visited but Did Not Make Any Transactions/Solution.sql

@@ -1,10 +1,7 @@
-SELECT
-    customer_id,
-    COUNT(*) AS count_no_trans
-FROM Visits
-WHERE
-    visit_id NOT IN (
-        SELECT visit_id
-        FROM Transactions
-    )
-GROUP BY customer_id;
+# Write your MySQL query statement below
+SELECT customer_id, COUNT(1) AS count_no_trans
+FROM
+    Visits
+    LEFT JOIN Transactions USING (visit_id)
+WHERE amount IS NULL
+GROUP BY 1;

solution/1600-1699/1683.Invalid Tweets/README.md

@@ -55,10 +55,13 @@ Tweets 表：
 
 <!-- 这里可写通用的实现逻辑 -->
 
--   `CHAR_LENGTH(str)`: 中文、数字、字母都是 1 字节
--   `LENGTH(str)`:
-    -   utf8: 中文 3 字节，数字、字母 1 字节
-    -   gbk: 中文 2 字节，数字、字母 1 字节
+**方法一：使用 `CHAR_LENGTH` 函数**
+
+`CHAR_LENGTH()` 函数返回字符串的长度，其中中文、数字、字母都是 $1$ 字节。
+
+`LENGTH()` 函数返回字符串的长度，其中 utf8 编码下，中文 $3$ 字节，数字、字母 $1$ 字节；gbk 编码下，中文 $2$ 字节，数字、字母 $1$ 字节。
+
+对于本题，我们直接用 `CHAR_LENGTH` 函数获取字符串长度，筛选出长度大于 $15$ 的推文 ID 即可。
 
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solution/1600-1699/1683.Invalid Tweets/README_EN.md

@@ -50,6 +50,14 @@ Tweet 2 has length = 32. It is an invalid tweet.
 
 ## Solutions
 
+**Solution 1: Using `CHAR_LENGTH` Function**
+
+The `CHAR_LENGTH()` function returns the length of a string, where Chinese characters, numbers, and letters are all counted as $1$ byte.
+
+The `LENGTH()` function returns the length of a string, where under utf8 encoding, Chinese characters are counted as $3$ bytes, while numbers and letters are counted as $1$ byte; under gbk encoding, Chinese characters are counted as $2$ bytes, while numbers and letters are counted as $1$ byte.
+
+For this problem, we can directly use the `CHAR_LENGTH` function to get the length of the string, and filter out the tweet IDs with a length greater than $15$.
+
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 ### **SQL**

solution/1700-1799/1757.Recyclable and Low Fat Products/README.md

@@ -59,6 +59,10 @@ Products 表：
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：条件筛选**
+
+我们直接筛选出 `low_fats` 为 `Y` 且 `recyclable` 为 `Y` 的产品编号即可。
+
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 ### **SQL**

solution/1700-1799/1757.Recyclable and Low Fat Products/README_EN.md

@@ -53,6 +53,10 @@ Products table:
 
 ## Solutions
 
+**Solution 1: Conditional Filtering**
+
+We can directly filter the product IDs where `low_fats` is `Y` and `recyclable` is `Y`.
+
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 ### **SQL**