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发表于 更新于 分类于 Valine：
本文字数： 1.4k 阅读时长 ≈ 1 分钟

You are given coins of different denominations and a total amount of money amount. Write a function to compute the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.

You may assume that you have an infinite number of each kind of coin.

Example 1:

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Input: coins = [1,2,5], amount = 11
Output: 3
Explanation: 11 = 5 + 5 + 1

Example 2:

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Input: coins = [2], amount = 3
Output: -1

Example 3:

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Input: coins = [1], amount = 0
Output: 0

Example 4:

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Input: coins = [1], amount = 1
Output: 1

Example 5:

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Input: coins = [1], amount = 2
Output: 2

Constraints:

  • 1 <= coins.length <= 12
  • 1 <= coins[i] <= 231 - 1
  • 0 <= amount <= 104
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发表于 更新于 分类于 Valine：
本文字数： 2.4k 阅读时长 ≈ 2 分钟

Given a non-empty array nums containing only positive integers, find if the array can be partitioned into two subsets such that the sum of elements in both subsets is equal.

Example 1:

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Input: nums = [1,5,11,5]
Output: true
Explanation: The array can be partitioned as [1, 5, 5] and [11].

Example 2:

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Input: nums = [1,2,3,5]
Output: false
Explanation: The array cannot be partitioned into equal sum subsets.

Constraints:

  • 1 <= nums.length <= 200
  • 1 <= nums[i] <= 100
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发表于 更新于 分类于 Valine：
本文字数： 3.5k 阅读时长 ≈ 3 分钟

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Example 1:

img

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Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
Output: 6
Explanation: The LCA of nodes 2 and 8 is 6.

Example 2:

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Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
Output: 2
Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

Example 3:

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Input: root = [2,1], p = 2, q = 1
Output: 2

Constraints:

  • The number of nodes in the tree is in the range [2, 105].
  • -109 <= Node.val <= 109
  • All Node.val are unique.
  • p != q
  • p and q will exist in the BST.
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发表于 更新于 分类于 Valine：
本文字数： 1.5k 阅读时长 ≈ 1 分钟

Given a list of non-negative integers nums, arrange them such that they form the largest number.

Note: The result may be very large, so you need to return a string instead of an integer.

Example 1:

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Input: nums = [10,2]
Output: "210"

Example 2:

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Input: nums = [3,30,34,5,9]
Output: "9534330"

Example 3:

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Input: nums = [1]
Output: "1"

Example 4:

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Input: nums = [10]
Output: "10"

Constraints:

  • 1 <= nums.length <= 100
  • 0 <= nums[i] <= 109
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发表于 更新于 分类于 Valine：
本文字数： 1.8k 阅读时长 ≈ 2 分钟

Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.

Example 1:

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Input: [3,2,1,5,6,4] and k = 2
Output: 5

Example 2:

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Input: [3,2,3,1,2,4,5,5,6] and k = 4
Output: 4

Note:
You may assume k is always valid, 1 ≤ k ≤ array’s length.

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发表于 更新于 分类于 Valine：
本文字数： 3.6k 阅读时长 ≈ 3 分钟

able: Department

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+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| revenue       | int     |
| month         | varchar |
+---------------+---------+
(id, month) is the primary key of this table.
The table has information about the revenue of each department per month.
The month has values in ["Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec"].

Write an SQL query to reformat the table such that there is a department id column and a revenue column for each month.

The query result format is in the following example:

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Department table:
+------+---------+-------+
| id   | revenue | month |
+------+---------+-------+
| 1    | 8000    | Jan   |
| 2    | 9000    | Jan   |
| 3    | 10000   | Feb   |
| 1    | 7000    | Feb   |
| 1    | 6000    | Mar   |
+------+---------+-------+

Result table:
+------+-------------+-------------+-------------+-----+-------------+
| id   | Jan_Revenue | Feb_Revenue | Mar_Revenue | ... | Dec_Revenue |
+------+-------------+-------------+-------------+-----+-------------+
| 1    | 8000        | 7000        | 6000        | ... | null        |
| 2    | 9000        | null        | null        | ... | null        |
| 3    | null        | 10000       | null        | ... | null        |
+------+-------------+-------------+-------------+-----+-------------+

Note that the result table has 13 columns (1 for the department id + 12 for the months).
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发表于 分类于 Valine：
本文字数： 5.2k 阅读时长 ≈ 5 分钟

这几天在做b站的一个爬虫项目.

先放api. 感觉爬虫阶段最难的地方就在于通过抓包或者其他手段来找api了, 需要与被爬网站斗智斗勇. 但是说实话b站对爬虫还算比较友好的了. 像京东什么的都需要直接上selenium了.

https://api.bilibili.com/x/web-interface/view?aid={}

aid应该是递增的. 所以从某个值开始遍历就可以了. 中间可能会碰到一些已经被删除了的视频, 这时返回的json文件中是没有”data”这个标签的, 直接跳过即可.

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发表于 分类于 Valine：
本文字数： 899 阅读时长 ≈ 1 分钟

在WSL上安装好后MySQL, 但是WSL并没有图形界面. 每次用命令行不方便. 于是想用win下面的MySQL workbench连接wsl上的MySQL.

在网上查了很多解决方案, 最后都没有解决问题. 踩了很多坑. 特此记录下解决问题的步骤, 供网友参考.

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发表于 更新于 分类于 Valine：
本文字数： 1k 阅读时长 ≈ 1 分钟

Mary is a teacher in a middle school and she has a table seat storing students’ names and their corresponding seat ids.

The column id is continuous increment.

Mary wants to change seats for the adjacent students.

Can you write a SQL query to output the result for Mary?

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+---------+---------+
|    id   | student |
+---------+---------+
|    1    | Abbot   |
|    2    | Doris   |
|    3    | Emerson |
|    4    | Green   |
|    5    | Jeames  |
+---------+---------+

For the sample input, the output is:

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+---------+---------+
|    id   | student |
+---------+---------+
|    1    | Doris   |
|    2    | Abbot   |
|    3    | Green   |
|    4    | Emerson |
|    5    | Jeames  |
+---------+---------+

Note:
If the number of students is odd, there is no need to change the last one’s seat.

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发表于 更新于 分类于 Valine：
本文字数： 1.4k 阅读时长 ≈ 1 分钟

The Employee table holds all employees. Every employee has an Id, and there is also a column for the department Id.

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+----+-------+--------+--------------+
| Id | Name  | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1  | Joe   | 85000  | 1            |
| 2  | Henry | 80000  | 2            |
| 3  | Sam   | 60000  | 2            |
| 4  | Max   | 90000  | 1            |
| 5  | Janet | 69000  | 1            |
| 6  | Randy | 85000  | 1            |
| 7  | Will  | 70000  | 1            |
+----+-------+--------+--------------+

The Department table holds all departments of the company.

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+----+----------+
| Id | Name     |
+----+----------+
| 1  | IT       |
| 2  | Sales    |
+----+----------+

Write a SQL query to find employees who earn the top three salaries in each of the department. For the above tables, your SQL query should return the following rows (order of rows does not matter).

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+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Max      | 90000  |
| IT         | Randy    | 85000  |
| IT         | Joe      | 85000  |
| IT         | Will     | 70000  |
| Sales      | Henry    | 80000  |
| Sales      | Sam      | 60000  |
+------------+----------+--------+

Explanation:

In IT department, Max earns the highest salary, both Randy and Joe earn the second highest salary, and Will earns the third highest salary. There are only two employees in the Sales department, Henry earns the highest salary while Sam earns the second highest salary.

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