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Commit 0c24e5b

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feat: update solutions to lc problems: No.0197,0584,1581,1683,1757 (doocs#1828)
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‎solution/0100-0199/0197.Rising Temperature/README.md‎

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@@ -61,6 +61,10 @@ Weather 表:</code>
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:自连接 + DATEDIFF/SUBDATE 函数**
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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**
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```
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```sql
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SELECT
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w2.idAS Id
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# Write your MySQL query statement below
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SELECTw1.id
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FROM
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weather AS w1
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JOIN weather AS w2 ON DATE_ADD( w1.recordDate, INTERVAL 1 DAY) = w2.recordDate
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WHERE
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w1.temperature < w2.temperature
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Weather AS w1
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JOIN Weather AS w2
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ON SUBDATE(w1.recordDate, 1) = w2.recordDate AND w1.temperature > w2.temperature;
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```
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‎solution/0100-0199/0197.Rising Temperature/README_EN.md‎

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@@ -54,6 +54,10 @@ In 2015年01月04日, the temperature was higher than the previous day (20 -&gt; 30).
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## Solutions
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**Solution 1: Self-Join + DATEDIFF/SUBDATE Function**
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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**
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```
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```sql
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SELECT
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w2.idAS Id
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# Write your MySQL query statement below
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SELECTw1.id
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FROM
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weather AS w1
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JOIN weather AS w2 ON DATE_ADD( w1.recordDate, INTERVAL 1 DAY) = w2.recordDate
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WHERE
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w1.temperature < w2.temperature
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Weather AS w1
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JOIN Weather AS w2
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ON SUBDATE(w1.recordDate, 1) = w2.recordDate AND w1.temperature > w2.temperature;
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```
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‎solution/0100-0199/0197.Rising Temperature/Solution.sql‎

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

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

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@@ -56,6 +56,10 @@ Customer 表:
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:条件过滤**
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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

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## Solutions
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**Solution 1: Conditional Filtering**
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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

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@@ -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。
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:子查询 + 分组统计**
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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。
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**方法二:左连接 + 分组统计**
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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**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```sql
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SELECT
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customer_id,
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COUNT(*) AS count_no_trans
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# Write your MySQL query statement below
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SELECT customer_id, COUNT(1) AS count_no_trans
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FROM Visits
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WHERE
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visit_id NOT IN (
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SELECT visit_id
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FROM Transactions
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)
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GROUP BY customer_id;
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WHERE visit_id NOT IN (SELECT visit_id FROM Transactions)
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GROUP BY 1;
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```
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```sql
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# Write your MySQL query statement below
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SELECT customer_id, COUNT(1) AS count_no_trans
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FROM
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VisitsAS v
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LEFT JOIN Transactions AS t ONv.visit_id=t.visit_id
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Visits
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LEFT JOIN Transactions USING (visit_id)
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WHERE amount IS NULL
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GROUP BY customer_id;
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GROUP BY 1;
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```
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‎solution/1500-1599/1581.Customer Who Visited but Did Not Make Any Transactions/README_EN.md‎

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## Solutions
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**Solution 1: Subquery + Grouping**
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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.
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**Solution 2: Left Join + Grouping**
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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**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```sql
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SELECT
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customer_id,
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COUNT(*) AS count_no_trans
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# Write your MySQL query statement below
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SELECT customer_id, COUNT(1) AS count_no_trans
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FROM Visits
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WHERE
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visit_id NOT IN (
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SELECT visit_id
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FROM Transactions
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)
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GROUP BY customer_id;
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WHERE visit_id NOT IN (SELECT visit_id FROM Transactions)
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GROUP BY 1;
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```
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```sql
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# Write your MySQL query statement below
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SELECT customer_id, COUNT(1) AS count_no_trans
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FROM
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VisitsAS v
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LEFT JOIN Transactions AS t ONv.visit_id=t.visit_id
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Visits
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LEFT JOIN Transactions USING (visit_id)
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WHERE amount IS NULL
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GROUP BY customer_id;
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GROUP BY 1;
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```
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SELECT
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customer_id,
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COUNT(*) AS count_no_trans
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FROM Visits
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WHERE
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visit_id NOT IN (
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SELECT visit_id
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FROM Transactions
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)
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GROUP BY customer_id;
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# Write your MySQL query statement below
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SELECT customer_id, COUNT(1) AS count_no_trans
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FROM
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Visits
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LEFT JOIN Transactions USING (visit_id)
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WHERE amount IS NULL
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GROUP BY 1;

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

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- `CHAR_LENGTH(str)`: 中文、数字、字母都是 1 字节
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- `LENGTH(str)`:
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- utf8: 中文 3 字节,数字、字母 1 字节
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- gbk: 中文 2 字节,数字、字母 1 字节
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**方法一:使用 `CHAR_LENGTH` 函数**
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`CHAR_LENGTH()` 函数返回字符串的长度,其中中文、数字、字母都是 1ドル$ 字节。
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`LENGTH()` 函数返回字符串的长度,其中 utf8 编码下,中文 3ドル$ 字节,数字、字母 1ドル$ 字节;gbk 编码下,中文 2ドル$ 字节,数字、字母 1ドル$ 字节。
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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ドル$ 的推文 ID 即可。
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‎solution/1600-1699/1683.Invalid Tweets/README_EN.md‎

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## Solutions
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**Solution 1: Using `CHAR_LENGTH` Function**
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The `CHAR_LENGTH()` function returns the length of a string, where Chinese characters, numbers, and letters are all counted as 1ドル$ byte.
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The `LENGTH()` function returns the length of a string, where under utf8 encoding, Chinese characters are counted as 3ドル$ bytes, while numbers and letters are counted as 1ドル$ byte; under gbk encoding, Chinese characters are counted as 2ドル$ bytes, while numbers and letters are counted as 1ドル$ byte.
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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ドル$.
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### **SQL**

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