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Commit fffbcb0

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feat: add solutions to lc problem: No.3236 (doocs#3347)
1 parent 62ac0a2 commit fffbcb0

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‎solution/1000-1099/1003.Check If Word Is Valid After Substitutions/README.md‎

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### 方法一:栈
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我们观察题目中的操作,可以发现,每一次都会在字符串的任意位置插入字符串 `"abc"`,所以每次插入操作之后,字符串的长度都会增加 3ドル$。如果字符串 $s$ 有效,那么它的长度一定是 3ドル$ 的倍数。因此,我们先对字符串 $s$ 的长度进行判断,如果不是 3ドル$ 的倍数,那么 $s$ 一定无效,可以直接返回 `false`。
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我们观察题目中的操作,可以发现,每一次都会在字符串的任意位置插入字符串 $\textit{"abc"}$,所以每次插入操作之后,字符串的长度都会增加 3ドル$。如果字符串 $s$ 有效,那么它的长度一定是 3ドル$ 的倍数。因此,我们先对字符串 $s$ 的长度进行判断,如果不是 3ドル$ 的倍数,那么 $s$ 一定无效,可以直接返回 $\textit{false}$。
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接下来我们遍历字符串 $s$ 的每个字符 $c,ドル我们先将字符 $c$ 压入栈 $t$ 中。如果此时栈 $t$ 的长度大于等于 3ドル,ドル并且栈顶的三个元素组成了字符串 `"abc"`,那么我们就将栈顶的三个元素弹出。然后继续遍历字符串 $s$ 的下一个字符。
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接下来我们遍历字符串 $s$ 的每个字符 $c,ドル我们先将字符 $c$ 压入栈 $t$ 中。如果此时栈 $t$ 的长度大于等于 3ドル,ドル并且栈顶的三个元素组成了字符串 $\textit{"abc"}$,那么我们就将栈顶的三个元素弹出。然后继续遍历字符串 $s$ 的下一个字符。
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遍历结束之后,如果栈 $t$ 为空,那么说明字符串 $s$ 有效,返回 `true`,否则返回 `false`。
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遍历结束之后,如果栈 $t$ 为空,那么说明字符串 $s$ 有效,返回 $\textit{true}$;否则,返回 $\textit{false}$。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 是字符串 $s$ 的长度。
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‎solution/1000-1099/1003.Check If Word Is Valid After Substitutions/README_EN.md‎

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### Solution 1: Stack
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If the string is valid, it's length must be the multiple of 3ドル$.
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We observe the operations in the problem and find that each time a string $\textit{"abc"}$ is inserted at any position in the string. Therefore, after each insertion operation, the length of the string increases by 3ドル$. If the string $s$ is valid, its length must be a multiple of 3ドル$. Thus, we first check the length of the string $s$. If it is not a multiple of 3ドル,ドル then $s$ must be invalid, and we can directly return $\textit{false}$.
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We traverse the string and push every character into the stack $t$. If the size of stack $t$ is greater than or equal to 3ドル$ and the top three elements of stack $t$ constitute the string `"abc"`, we pop the top three elements. Then we continue to traverse the next character of the string $s$.
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Next, we traverse each character $c$ in the string $s$. We first push the character $c$ onto the stack $t$. If the length of the stack $t$ is greater than or equal to 3ドル$, and the top three elements of the stack form the string $\textit{"abc"},ドル then we pop the top three elements from the stack. We then continue to traverse the next character in the string $s$.
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When the traversal is over, if the stack $t$ is empty, the string $s$ is valid, return `true`, otherwisereturn `false`.
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After the traversal, if the stack $t$ is empty, it means the string $s$ is valid, and we return $\textit{true}$; otherwise, we return $\textit{false}$.
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The time complexity is $O(n)$ and the space complexity is $O(n)$. Where $n$ is the length of the string $s$.
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The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $s$.
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<!-- tabs:start -->
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‎solution/3200-3299/3236.CEO Subordinate Hierarchy/README.md‎

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@@ -105,14 +105,56 @@ manager_id 是 employee_id 对应员工的经理。首席执行官的 manager_id
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<!-- solution:start -->
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### 方法一
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### 方法一:递归 CTE + 连接
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首先,我们使用递归 CTE 计算出每个员工的层级,其中 CEO 的层级为 0,将 `employee_id`、`employee_name`、`hierarchy_level`、`manager_id` 和 `salary` 保存到临时表 `T` 中。
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然后,我们查询出 CEO 的薪资,将其保存到临时表 `P` 中。
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最后,我们连接 `T` 和 `P` 表,计算出每个下属的薪资差异,并按照 `hierarchy_level` 和 `subordinate_id` 进行排序。
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<!-- tabs:start -->
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#### MySQL
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```sql
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# Write your MySQL query statement below
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WITH RECURSIVE
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T AS (
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SELECT
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employee_id,
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employee_name,
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0 AS hierarchy_level,
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manager_id,
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salary
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FROM Employees
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WHERE manager_id IS NULL
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UNION ALL
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SELECT
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e.employee_id,
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e.employee_name,
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hierarchy_level + 1 AS hierarchy_level,
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e.manager_id,
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e.salary
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FROM
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T t
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JOIN Employees e ON t.employee_id = e.manager_id
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),
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P AS (
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SELECT salary
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FROM Employees
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WHERE manager_id IS NULL
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)
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SELECT
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employee_id subordinate_id,
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employee_name subordinate_name,
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hierarchy_level,
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t.salary - p.salary salary_difference
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FROM
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T t
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JOIN P p
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WHERE hierarchy_level != 0
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ORDER BY 3, 1;
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```
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<!-- tabs:end -->

‎solution/3200-3299/3236.CEO Subordinate Hierarchy/README_EN.md‎

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@@ -106,14 +106,56 @@ manager_id is the employee_id of the employee&#39;s manager. The CEO has a NULL
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Recursive CTE + Join
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First, we use a recursive CTE to calculate the hierarchy level of each employee, where the CEO's level is 0ドル$. We save `employee_id`, `employee_name`, `hierarchy_level`, `manager_id`, and `salary` into a temporary table `T`.
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Then, we query the CEO's salary and save it into a temporary table `P`.
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Finally, we join tables `T` and `P` to calculate the salary difference for each subordinate, and sort by `hierarchy_level` and `subordinate_id`.
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<!-- tabs:start -->
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#### MySQL
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```sql
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# Write your MySQL query statement below
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WITH RECURSIVE
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T AS (
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SELECT
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employee_id,
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employee_name,
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0 AS hierarchy_level,
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manager_id,
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salary
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FROM Employees
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WHERE manager_id IS NULL
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UNION ALL
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SELECT
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e.employee_id,
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e.employee_name,
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hierarchy_level + 1 AS hierarchy_level,
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e.manager_id,
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e.salary
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FROM
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T t
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JOIN Employees e ON t.employee_id = e.manager_id
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),
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P AS (
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SELECT salary
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FROM Employees
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WHERE manager_id IS NULL
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)
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SELECT
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employee_id subordinate_id,
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employee_name subordinate_name,
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hierarchy_level,
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t.salary - p.salary salary_difference
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FROM
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T t
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JOIN P p
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WHERE hierarchy_level != 0
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ORDER BY 3, 1;
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```
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<!-- tabs:end -->
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# Write your MySQL query statement below
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WITH RECURSIVE
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T AS (
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SELECT
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employee_id,
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employee_name,
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0 AS hierarchy_level,
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manager_id,
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salary
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FROM Employees
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WHERE manager_id IS NULL
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UNION ALL
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SELECT
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e.employee_id,
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e.employee_name,
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hierarchy_level + 1 AS hierarchy_level,
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e.manager_id,
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e.salary
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FROM
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T t
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JOIN Employees e ON t.employee_id = e.manager_id
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),
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P AS (
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SELECT salary
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FROM Employees
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WHERE manager_id IS NULL
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)
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SELECT
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employee_id subordinate_id,
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employee_name subordinate_name,
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hierarchy_level,
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t.salary - p.salary salary_difference
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FROM
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T t
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JOIN P p
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WHERE hierarchy_level != 0
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ORDER BY 3, 1;

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