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83.删除排序链表中的重复元素、108.将有序数组转换为二叉搜索树、181.超过经理收入的员工、182.查找重复的电子邮箱

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游由 2022-08-10 23:10:06 +08:00
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2
rust/src/implements/mod.rs
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@ -39,6 +39,7 @@ mod q0070;
mod q0078;
mod q0080;
mod q0081;
mod q0083;
mod q0086;
mod q0087;
mod q0088;
@ -49,6 +50,7 @@ mod q0100;
mod q0101;
mod q0102;
mod q0104;
mod q0108;
mod q0112;
mod q0110;
mod q0111;

65
rust/src/implements/q0083.rs Normal file
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@ -0,0 +1,65 @@
use crate::Solution;
use crate::structure::ListNode;
impl Solution {
/// [83.删除排序链表中的重复元素](https://leetcode-cn.com/problems/remove-duplicates-from-sorted-list/)
///
/// 2022-08-10 21:10:32
///
/// 给定一个已排序的链表的头 `head` ， _删除所有重复的元素，使每个元素只出现一次_ 。返回 _已排序的链表_ 。
///
/// + **示例 1：**
/// + ![](https://assets.leetcode.com/uploads/2021/01/04/list1.jpg)```
/// + **输入：**head = \[1,1,2\]
/// + **输出：**\[1,2\]
/// + **示例 2：**
/// + ![](https://assets.leetcode.com/uploads/2021/01/04/list2.jpg)```
/// + **输入：**head = \[1,1,2,3,3\]
/// + **输出：**\[1,2,3\]
/// + **提示：**
/// * 链表中节点数目在范围 `[0, 300]` 内
/// * `-100 <= Node.val <= 100`
/// * 题目数据保证链表已经按升序 **排列**
/// + Related Topics
/// * 链表
pub fn delete_duplicates(mut head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
/*let mut vec = Vec::<i32>::new();
while let Some(n) = head {
if Some(&n.val) != vec.last() {
vec.push(n.val);
}
head = n.next;
}
let mut node = None;
for val in vec.into_iter().rev() {
node = Some(Box::new(ListNode { val, next: node }));
}
node*/
if head.is_none() {
return None
}
let mut node = head.as_mut().unwrap();
let mut pre = node.val;
while let Some(mut next) = node.next.take() {
if pre == next.val { //去掉当前节点
node.next = next.next;
} else { // 还原
pre = next.val;
node.next = Some(next);
node = node.next.as_mut().unwrap();
}
}
head
}
}
#[cfg(test)]
mod test {
use crate::Solution;
use crate::structure::ListNode;
#[test]
fn test_q0083() {
assert_eq!(Solution::delete_duplicates(Some(Box::new(ListNode { val: 1, next: Some(Box::new(ListNode { val: 1, next: Some(Box::new(ListNode { val: 2, next: None })) })) }))), Some(Box::new(ListNode { val: 1, next: Some(Box::new(ListNode { val: 2, next: None })) })));
}
}

54
rust/src/implements/q0108.rs Normal file
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@ -0,0 +1,54 @@
use crate::Solution;
use crate::structure::TreeNode;
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
/// [108.将有序数组转换为二叉搜索树](https://leetcode-cn.com/problems/convert-sorted-array-to-binary-search-tree/)
///
/// 2022-08-10 21:43:44
///
/// 给你一个整数数组 `nums` ，其中元素已经按 **升序** 排列，请你将其转换为一棵 **高度平衡** 二叉搜索树。
///
/// **高度平衡** 二叉树是一棵满足「每个节点的左右两个子树的高度差的绝对值不超过 1 」的二叉树。
///
/// + **示例 1：**
/// + ![](https://assets.leetcode.com/uploads/2021/02/18/btree1.jpg)
/// + **输入：** nums = \[-10,-3,0,5,9\]
/// + **输出：** \[0,-3,9,-10,null,5\]
/// + **解释：** \[0,-10,5,null,-3,null,9\] 也将被视为正确答案：
/// + ![](https://assets.leetcode.com/uploads/2021/02/18/btree2.jpg)
/// + **示例 2：**
/// + ![](https://assets.leetcode.com/uploads/2021/02/18/btree.jpg)
/// + **输入：** nums = \[1,3\]
/// + **输出：** \[3,1\]
/// + **解释：** \[1,null,3\] 和 \[3,1\] 都是高度平衡二叉搜索树。
/// + **提示：**
/// * `1 <= nums.length <= 104`
/// * `-104 <= nums[i] <= 104`
/// * `nums` 按 **严格递增** 顺序排列
/// + Related Topics
/// * 树
/// * 二叉搜索树
/// * 数组
/// * 分治
/// * 二叉树
pub fn sorted_array_to_bst(nums: Vec<i32>) -> Option<Rc<RefCell<TreeNode>>> {
match nums.len() {
0 => None,
l => Some(Rc::new(RefCell::new(TreeNode { val: nums[l / 2], left: Self::sorted_array_to_bst(nums[..l / 2].to_vec()), right: Self::sorted_array_to_bst(nums[l / 2 + 1..].to_vec()) })))
}
}
}
#[cfg(test)]
mod test {
use crate::Solution;
#[test]
fn test_q0108() {
println!("{:?}", Solution::sorted_array_to_bst(vec![1, 3]));
println!("{:?}", Solution::sorted_array_to_bst(vec![-10, -3, 0, 5, 9]));
}
}

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sql/q0181.md Normal file
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# [181.超过经理收入的员工](https://leetcode-cn.com/problems/employees-earning-more-than-their-managers/)
2022-08-10 22:05:28
## SQL架构
```mysql
Create table If Not Exists Employee
(
id int,
name varchar(255),
salary int,
managerId int
);
Truncate table Employee;
insert into Employee (id, name, salary, managerId)
values ('1', 'Joe', '70000', '3');
insert into Employee (id, name, salary, managerId)
values ('2', 'Henry', '80000', '4');
insert into Employee (id, name, salary, managerId)
values ('3', 'Sam', '60000', 'None');
insert into Employee (id, name, salary, managerId)
values ('4', 'Max', '90000', 'None');
```
## 题
表：`Employee`
```
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| id | int |
| name | varchar |
| salary | int |
| managerId | int |
+-------------+---------+
Id是该表的主键。
该表的每一行都表示雇员的ID、姓名、工资和经理的ID。
```
编写一个SQL查询来查找收入比经理高的员工。
以 **任意顺序** 返回结果表。
查询结果格式如下所示。
**示例 1:**
```
**输入:**
Employee 表:
+----+-------+--------+-----------+
| id | name | salary | managerId |
+----+-------+--------+-----------+
| 1 | Joe | 70000 | 3 |
| 2 | Henry | 80000 | 4 |
| 3 | Sam | 60000 | Null |
| 4 | Max | 90000 | Null |
+----+-------+--------+-----------+
**输出:**
+----------+
| Employee |
+----------+
| Joe |
+----------+
**解释:** Joe 是唯一挣得比经理多的雇员。
```
Related Topics
* 数据库
## 解
```mysql
SELECT a.Name AS 'Employee'
FROM Employee AS a,
Employee AS b
WHERE a.ManagerId = b.Id
AND a.Salary > b.Salary;
```
```mysql
SELECT a.Name AS 'Employee'
FROM Employee a,
Employee b
WHERE a.ManagerId = b.Id
AND a.Salary > b.Salary;
```
```mysql
SELECT a.NAME AS Employee
FROM Employee AS a
JOIN Employee AS b
ON a.ManagerId = b.Id
AND a.Salary > b.Salary;
```
```mysql
SELECT a.NAME AS Employee
FROM Employee a
JOIN Employee b
ON a.ManagerId = b.Id
AND a.Salary > b.Salary;
```
不加`AS`的快一点好像。

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sql/q0182.md Normal file
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@ -0,0 +1,74 @@
# [182.查找重复的电子邮箱](https://leetcode-cn.com/problems/duplicate-emails/)
2022-08-10 22:24:58
## SQL架构
```mysql
Create table If Not Exists Person
(
id int,
email varchar(255)
);
Truncate table Person;
insert into Person (id, email)
values ('1', 'a@b.com');
insert into Person (id, email)
values ('2', 'c@d.com');
insert into Person (id, email)
values ('3', 'a@b.com');
```
## 题目
编写一个 SQL 查询，查找 `Person` 表中所有重复的电子邮箱。
**示例：**
```
+----+---------+
| Id | Email |
+----+---------+
| 1 | a@b.com |
| 2 | c@d.com |
| 3 | a@b.com |
+----+---------+
```
根据以上输入，你的查询应返回以下结果：
```
+---------+
| Email |
+---------+
| a@b.com |
+---------+
```
**说明：**所有电子邮箱都是小写字母。
Related Topics
* 数据库
## 解
```mysql
select Email
from (select Email, count(Email) as num
from Person
group by Email) as statistic
where num > 1;
```
```mysql
select Email
from Person
group by Email
having count(Email) > 1;
```