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feat: update solutions to lc problems: No.0534,0608,1264,1965 #1791

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yanglbme merged 2 commits into main from dev
Oct 12, 2023
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33 changes: 31 additions & 2 deletions solution/0500-0599/0534.Game Play Analysis III/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -66,9 +66,13 @@ Activity table:

<!-- 这里可写通用的实现逻辑 -->

**方法一:SUM() OVER() 窗口函数**
**方法一:使用窗口函数**

我们可以使用 `SUM() OVER()` 窗口函数来计算每个玩家到目前为止玩了多少游戏。在 `OVER()` 子句中,我们使用 `PARTITION BY` 子句将玩家分组,然后使用 `ORDER BY` 子句按日期排序。
我们可以使用窗口函数 `SUM() OVER()`,按照 `player_id` 分组,按照 `event_date` 排序,计算每个用户截止到当前日期的游戏总数。

**方法二:使用自连接 + 分组**

我们也可以使用自连接,将 `Activity` 表自连接,连接条件为 `t1.player_id = t2.player_id AND t1.event_date >= t2.event_date`,然后按照 `t1.player_id` 和 `t1.event_date` 分组,累计 `t2.games_played`,得到每个用户截止到当前日期的游戏总数。

<!-- tabs:start -->

Expand All @@ -86,4 +90,29 @@ SELECT
FROM Activity;
```

```sql
# Write your MySQL query statement below
SELECT
t1.player_id,
t1.event_date,
sum(t2.games_played) AS games_played_so_far
FROM
Activity AS t1,
Activity AS t2
WHERE t1.player_id = t2.player_id AND t1.event_date >= t2.event_date
GROUP BY 1, 2;
```

```sql
# Write your MySQL query statement below
SELECT
t1.player_id,
t1.event_date,
sum(t2.games_played) AS games_played_so_far
FROM
Activity AS t1
CROSS JOIN Activity AS t2 ON t1.player_id = t2.player_id AND t1.event_date >= t2.event_date
GROUP BY 1, 2;
```

<!-- tabs:end -->
33 changes: 33 additions & 0 deletions solution/0500-0599/0534.Game Play Analysis III/README_EN.md
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Expand Up @@ -61,6 +61,14 @@ Note that for each player we only care about the days when the player logged in.

## Solutions

**Solution 1: Window Function**

We can use the window function `SUM() OVER()` to group by `player_id`, sort by `event_date`, and calculate the total number of games played by each user up to the current date.

**Solution 2: Self-Join + Group By**

We can also use a self-join to join the `Activity` table with itself on the condition of `t1.player_id = t2.player_id AND t1.event_date >= t2.event_date`, and then group by `t1.player_id` and `t1.event_date`, and calculate the cumulative sum of `t2.games_played`. This will give us the total number of games played by each user up to the current date.

<!-- tabs:start -->

### **SQL**
Expand All @@ -77,4 +85,29 @@ SELECT
FROM Activity;
```

```sql
# Write your MySQL query statement below
SELECT
t1.player_id,
t1.event_date,
sum(t2.games_played) AS games_played_so_far
FROM
Activity AS t1,
Activity AS t2
WHERE t1.player_id = t2.player_id AND t1.event_date >= t2.event_date
GROUP BY 1, 2;
```

```sql
# Write your MySQL query statement below
SELECT
t1.player_id,
t1.event_date,
sum(t2.games_played) AS games_played_so_far
FROM
Activity AS t1
CROSS JOIN Activity AS t2 ON t1.player_id = t2.player_id AND t1.event_date >= t2.event_date
GROUP BY 1, 2;
```

<!-- tabs:end -->
8 changes: 8 additions & 0 deletions solution/0600-0699/0608.Tree Node/README.md
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Expand Up @@ -91,6 +91,14 @@ Tree table:

<!-- 这里可写通用的实现逻辑 -->

**方法一:条件判断 + 子查询**

我们可以使用 `CASE WHEN` 条件判断语句来判断每个节点的类型,具体地:

- 如果一个节点的 `p_id` 为 `NULL`,则该节点为根节点;
- 否则,如果一个节点是另一个节点的父节点(这里我们使用子查询来判断),则该节点为内部节点;
- 否则,该节点为叶子节点。

<!-- tabs:start -->

### **SQL**
Expand Down
8 changes: 8 additions & 0 deletions solution/0600-0699/0608.Tree Node/README_EN.md
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Expand Up @@ -86,6 +86,14 @@ Tree table:

## Solutions

**Solution 1: Conditional Statements + Subquery**

We can use the `CASE WHEN` conditional statement to determine the type of each node as follows:

- If a node's `p_id` is `NULL`, then it is a root node.
- Otherwise, if a node is the parent node of another node (we use a subquery to determine this), then it is an internal node.
- Otherwise, it is a leaf node.

<!-- tabs:start -->

### **SQL**
Expand Down
19 changes: 19 additions & 0 deletions solution/1200-1299/1264.Page Recommendations/README.md
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Expand Up @@ -96,10 +96,29 @@ Likes table:

<!-- 这里可写通用的实现逻辑 -->

**方法一:合并 + 等值连接 + 子查询**

我们先查出所有与 `user_id = 1` 的用户是朋友的用户,记录在 `T` 表中,然后再查出所有在 `T` 表中的用户喜欢的页面,最后排除掉 `user_id = 1` 喜欢的页面即可。

<!-- tabs:start -->

### **SQL**

```sql
# Write your MySQL query statement below
WITH
T AS (
SELECT user1_id AS user_id FROM Friendship WHERE user2_id = 1
UNION
SELECT user2_id AS user_id FROM Friendship WHERE user1_id = 1
)
SELECT DISTINCT page_id AS recommended_page
FROM
T
JOIN Likes USING (user_id)
WHERE page_id NOT IN (SELECT page_id FROM Likes WHERE user_id = 1);
```

```sql
# Write your MySQL query statement below
SELECT DISTINCT page_id AS recommended_page
Expand Down
19 changes: 19 additions & 0 deletions solution/1200-1299/1264.Page Recommendations/README_EN.md
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Expand Up @@ -90,10 +90,29 @@ Page 88 is not suggested because user 1 already likes it.

## Solutions

**Solution 1: Union + Equi-Join + Subquery**

First, we query all users who are friends with `user_id = 1` and record them in the `T` table. Then, we query all pages that users in the `T` table like, and finally exclude the pages that `user_id = 1` likes.

<!-- tabs:start -->

### **SQL**

```sql
# Write your MySQL query statement below
WITH
T AS (
SELECT user1_id AS user_id FROM Friendship WHERE user2_id = 1
UNION
SELECT user2_id AS user_id FROM Friendship WHERE user1_id = 1
)
SELECT DISTINCT page_id AS recommended_page
FROM
T
JOIN Likes USING (user_id)
WHERE page_id NOT IN (SELECT page_id FROM Likes WHERE user_id = 1);
```

```sql
# Write your MySQL query statement below
SELECT DISTINCT page_id AS recommended_page
Expand Down
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Original file line number Diff line number Diff line change
Expand Up @@ -83,29 +83,26 @@ Salaries table:

<!-- 这里可写通用的实现逻辑 -->

**方法一:子查询 + 合并**

我们可以先从 `Employees` 表中找出所有不在 `Salaries` 表中的 `employee_id`,再从 `Salaries` 表中找出所有不在 `Employees` 表中的 `employee_id`,最后将两个结果合并,然后按照 `employee_id` 排序即可。

<!-- tabs:start -->

### **SQL**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```sql
# Write your MySQL query statement below
SELECT employee_id
FROM Employees AS e
WHERE
e.employee_id NOT IN (
SELECT employee_id
FROM Salaries
)
FROM Employees
WHERE employee_id NOT IN (SELECT employee_id FROM Salaries)
UNION
SELECT employee_id
FROM Salaries AS s
WHERE
s.employee_id NOT IN (
SELECT employee_id
FROM Employees
)
ORDER BY employee_id;
FROM Salaries
WHERE employee_id NOT IN (SELECT employee_id FROM Employees)
ORDER BY 1;
```

<!-- tabs:end -->
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Expand Up @@ -81,27 +81,24 @@ The salary of employee 2 is missing.

## Solutions

**Solution 1: Subquery + Union**

We can first find all `employee_id` that are not in the `Salaries` table from the `Employees` table, and then find all `employee_id` that are not in the `Employees` table from the `Salaries` table. Finally, we can combine the two results using the `UNION` operator, and sort the result by `employee_id`.

<!-- tabs:start -->

### **SQL**

```sql
# Write your MySQL query statement below
SELECT employee_id
FROM Employees AS e
WHERE
e.employee_id NOT IN (
SELECT employee_id
FROM Salaries
)
FROM Employees
WHERE employee_id NOT IN (SELECT employee_id FROM Salaries)
UNION
SELECT employee_id
FROM Salaries AS s
WHERE
s.employee_id NOT IN (
SELECT employee_id
FROM Employees
)
ORDER BY employee_id;
FROM Salaries
WHERE employee_id NOT IN (SELECT employee_id FROM Employees)
ORDER BY 1;
```

<!-- tabs:end -->
View file Open in desktop
Original file line number Diff line number Diff line change
@@ -1,16 +1,9 @@
# Write your MySQL query statement below
SELECT employee_id
FROM Employees AS e
WHERE
e.employee_id NOT IN (
SELECT employee_id
FROM Salaries
)
FROM Employees
WHERE employee_id NOT IN (SELECT employee_id FROM Salaries)
UNION
SELECT employee_id
FROM Salaries AS s
WHERE
s.employee_id NOT IN (
SELECT employee_id
FROM Employees
)
ORDER BY employee_id;
FROM Salaries
WHERE employee_id NOT IN (SELECT employee_id FROM Employees)
ORDER BY 1;

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