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<!DOCTYPE html><html lang="en" data-theme="light"><head><meta charset="UTF-8"><meta http-equiv="X-UA-Compatible" content="IE=edge"><meta name="viewport" content="width=device-width, initial-scale=1.0, maximum-scale=1.0"><title>Hexo</title><meta name="author" content="John Doe"><meta name="copyright" content="John Doe"><meta name="format-detection" content="telephone=no"><meta name="theme-color" content="#ffffff"><meta property="og:type" content="website">
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})(window)</script><meta name="generator" content="Hexo 6.3.0"></head><body><div id="sidebar"><div id="menu-mask"></div><div id="sidebar-menus"><div class="avatar-img is-center"><img src="https://i.loli.net/2021/02/24/5O1day2nriDzjSu.png" onerror="onerror=null;src='/img/friend_404.gif'" alt="avatar"/></div><div class="sidebar-site-data site-data is-center"><a href="/archives/"><div class="headline">Articles</div><div class="length-num">15</div></a><a href="/tags/"><div class="headline">Tags</div><div class="length-num">0</div></a><a href="/categories/"><div class="headline">Categories</div><div class="length-num">0</div></a></div><hr/></div></div><div class="page" id="body-wrap"><header class="full_page" id="page-header"><nav id="nav"><span id="blog-info"><a href="/" title="Hexo"><span class="site-name">Hexo</span></a></span><div id="menus"><div id="toggle-menu"><a class="site-page" href="javascript:void(0);"><i class="fas fa-bars fa-fw"></i></a></div></div></nav><div id="site-info"><h1 id="site-title">Hexo</h1></div><div id="scroll-down"><i class="fas fa-angle-down scroll-down-effects"></i></div></header><main class="layout" id="content-inner"><div class="recent-posts" id="recent-posts"><div class="recent-post-item"><div class="recent-post-info no-cover"><a class="article-title" href="/2023/10/18/%E2%80%9C%E6%88%91%E7%9A%84%E7%AC%AC%E5%8D%81%E4%B8%89%E7%AF%87%E5%8D%9A%E5%AE%A2%E2%80%9D/" title="“我的第十三篇博客”">“我的第十三篇博客”</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">Created</span><time datetime="2023-10-18T01:18:53.000Z" title="Created 2023-10-18 09:18:53">2023-10-18</time></span></div><div class="content">题目：居中1.1 作者：姓名，大学，学院或实验室（居中）1.2 摘要：Abstract（居中）我们提出了一个命名为某某某的新的框架，是训练几个模型，每个模型具体是做什么的，通过什么来训练的，具体达到的效果是什么样的2.1 简介：1 Introduction（左对齐）深度学习的目标 到目前为止深度学习最引人注目的成果 这些成果是基于XXX 由于XXX的缺点和困难 本论文提出了一种XXX来解决这些困难提出的这个XXX框架 判别模型职责是什么 生成模型职责是什么该框架和传统意义上的框架相比有什么优势，具体是如何体现这些优势的2.2 相关工作当涉及深度生成模型时，我们可以将其想象成一个艺术家。以往的工作主要集中在那些可以用公式精确描述的模型上，就像是艺术家尝试复制一幅已经存在的画作一样。然而，复杂的模型，比如深度玻尔兹曼机，有时候会遇到一些困难，就好比艺术家试图复制一幅特别复杂的画作，但却只能依靠模糊的照片或者模糊的记忆。
因此，为了克服这些困难，人们开始探索一种新的方式。这种方式就像是让艺术家尝试创造一幅全新的作品，而不是简单地复制现有的作品。这样的模型并不依赖于精确的公式描述， ...</div></div></div><div class="recent-post-item"><div class="recent-post-info no-cover"><a class="article-title" href="/2023/10/11/%E2%80%9C%E6%88%91%E7%9A%84%E7%AC%AC%E5%8D%81%E4%BA%8C%E7%AF%87%E5%8D%9A%E5%AE%A2%E2%80%9D/" title="“我的第十二篇博客”">“我的第十二篇博客”</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">Created</span><time datetime="2023-10-11T07:41:03.000Z" title="Created 2023-10-11 15:41:03">2023-10-11</time></span></div><div class="content">进入markdown编辑模式运行 cmd 跳转到安装目录，输入 node_inject.exe 回车即可自动处理，然后输入 license-gen.exe 回车可得到序列号，接着输入任意邮箱和序列号进行激活，关闭重启Typora再输入序列号会激活成功。123456789101112C:\Users\Administrator>D:D:\>E:E:\>cd E:\Program Files\soft\typora\TyporaE:\Program Files\soft\typora\Typora> .\node_inject.exeextracting node_modules.asaradding hook.jsapplying patchpacking node_modules.asardone!E:\Program Files\soft\typora\Typora> .\license-gen.exeLicense for you: WE6KCN-YJ8L9H-ASK47A-PR2ANGE:\Program Files\soft\typora\Typo ...</div></div></div><div class="recent-post-item"><div class="recent-post-info no-cover"><a class="article-title" href="/2023/10/10/%E6%88%91%E7%9A%84%E7%AC%AC%E5%8D%81%E4%B8%80%E7%AF%87%E5%8D%9A%E5%AE%A2%E6%96%87%E7%AB%A0/" title="我的第十一篇博客文章">我的第十一篇博客文章</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">Created</span><time datetime="2023-10-10T00:58:04.000Z" title="Created 2023-10-10 08:58:04">2023-10-10</time></span></div><div class="content">GitHub 术语解释Repository：简称Repo，可以理解为“仓库”，我们的项目就存放在仓库之中。也就是说，如果我们想要建立项目，就得先建立仓库；有多个项目，就建立多个仓库。
Issues：可以理解为“问题”，举一个简单的例子，如果我们开源一个项目，如果别人看了我们的项目，并且发现了bug，或者感觉那个地方有待改进，他就可以给我们提出Issue，等我们把Issues解决之后，就可以把这些Issues关闭；反之，我们也可以给他人提出Issue。
Star：可以理解为“点赞”，当我们感觉某一个项目做的比较好之后，就可以为这个项目点赞，而且我们点赞过的项目，都会保存到我们的Star之中，方便我们随时查看。在 GitHub 之中，如果一个项目的点星数能够超百，那么说明这个项目已经很不错了。
Fork：可以理解为“拉分支”，如果我们对某一个项目比较感兴趣，并且想在此基础之上开发新的功能，这时我们就可以Fork这个项目，这表示复制一个完成相同的项目到我们的 GitHub 账号之中，而且独立于原项目。之后，我们就可以在自己复制的项目中进行开发了。
Pull Request：可以理解为“提交请 ...</div></div></div><div class="recent-post-item"><div class="recent-post-info no-cover"><a class="article-title" href="/2023/09/06/%E2%80%9C%E6%88%91%E7%9A%84%E7%AC%AC%E5%8D%81%E7%AF%87%E5%8D%9A%E5%AE%A2%E2%80%9D/" title="“我的第十篇博客”">“我的第十篇博客”</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">Created</span><time datetime="2023-09-06T07:08:09.000Z" title="Created 2023-09-06 15:08:09">2023-09-06</time></span></div><div class="content">买卖股票的最佳时机II问题描述：给你一个整数数组 prices ，其中 prices[i] 表示某支股票第 i 天的价格。在每一天，你可以决定是否购买和/或出售股票。你在任何时候 最多 只能持有 一股 股票。你也可以先购买，然后在 同一天 出售。返回 你能获得的 最大 利润 。解决方案：采用动态规划12345678910111213class Solution {public: int maxProfit(vector<int>& prices) { int n = prices.size(); int dp[n][2]; dp[0][0] = 0, dp[0][1] = -prices[0]; for (int i = 1; i < n; ++i) { dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] + prices[i]); dp[i][1] = max(dp[i - 1][1 ...</div></div></div><div class="recent-post-item"><div class="recent-post-info no-cover"><a class="article-title" href="/2023/09/06/%E2%80%9C%E6%88%91%E7%9A%84%E7%AC%AC%E4%B9%9D%E7%AF%87%E5%8D%9A%E5%AE%A2%E2%80%9D/" title="“我的第九篇博客”">“我的第九篇博客”</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">Created</span><time datetime="2023-09-06T06:01:06.000Z" title="Created 2023-09-06 14:01:06">2023-09-06</time></span></div><div class="content">买卖股票的最佳时机问题描述：给定一个数组 prices ，它的第 i 个元素 prices[i] 表示一支给定股票第 i 天的价格。
你只能选择 某一天 买入这只股票，并选择在 未来的某一个不同的日子 卖出该股票。设计一个算法来计算你所能获取的最大利润。
返回你可以从这笔交易中获取的最大利润。如果你不能获取任何利润，返回 0 。
解决方案：数学上形式化描述就是给定一个数组，然后寻找数组中差值最大的一组数，并计算差值返回，并且数组下标j需要满足大于i。123456789101112class Solution {public: int maxProfit(vector<int>& prices) { int n = (int)prices.size(), ans = 0; for (int i = 0; i < n; ++i){ for (int j = i + 1; j < n; ++j) { ans = max(ans, pri ...</div></div></div><div class="recent-post-item"><div class="recent-post-info no-cover"><a class="article-title" href="/2023/09/06/%E2%80%9C%E6%88%91%E7%9A%84%E7%AC%AC%E5%85%AB%E7%AF%87%E5%8D%9A%E5%AE%A2%E2%80%9D/" title="“我的第八篇博客”">“我的第八篇博客”</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">Created</span><time datetime="2023-09-06T01:16:51.000Z" title="Created 2023-09-06 09:16:51">2023-09-06</time></span></div><div class="content">轮转数组问题描述：给定一个整数数组 nums，将数组中的元素向右轮转 k 个位置，其中 k 是非负数。解决方案：需要用到取模运算，需要开辟一个新的数组进行元素归位，然后复制新数组到旧数组1234567891011class Solution {public: void rotate(vector<int>& nums, int k) { int n = nums.size(); vector<int> newArr(n); for (int i = 0; i < n; ++i) { newArr[(i + k) % n] = nums[i]; } nums.assign(newArr.begin(), newArr.end()); }};
</div></div></div><div class="recent-post-item"><div class="recent-post-info no-cover"><a class="article-title" href="/2023/09/05/%E2%80%9C%E6%88%91%E7%9A%84%E7%AC%AC%E4%B8%83%E7%AF%87%E5%8D%9A%E5%AE%A2%E2%80%9D/" title="“我的第七篇博客”">“我的第七篇博客”</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">Created</span><time datetime="2023-09-05T02:51:29.000Z" title="Created 2023-09-05 10:51:29">2023-09-05</time></span></div><div class="content">多数元素问题描述：给定一个大小为 n 的数组 nums ，返回其中的多数元素。多数元素是指在数组中出现次数 大于 ⌊ n/2 ⌋ 的元素。你可以假设数组是非空的，并且给定的数组总是存在多数元素。解决方案：我们使用哈希映射（HashMap）来存储每个元素以及出现的次数。对于哈希映射中的每个键值对，键表示一个元素，值表示该元素出现的次数。123456789101112131415class Solution {public: int majorityElement(vector<int>& nums) { unordered_map<int, int> counts; int majority = 0, cnt = 0; for (int num: nums) { ++counts[num]; if (counts[num] > cnt) { majority = num; ...</div></div></div><div class="recent-post-item"><div class="recent-post-info no-cover"><a class="article-title" href="/2023/09/05/%E2%80%9C%E6%88%91%E7%9A%84%E7%AC%AC%E5%85%AD%E7%AF%87%E5%8D%9A%E5%AE%A2%E2%80%9D/" title="“我的第六篇博客”">“我的第六篇博客”</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">Created</span><time datetime="2023-09-05T02:18:42.000Z" title="Created 2023-09-05 10:18:42">2023-09-05</time></span></div><div class="content">删除有序数组中的重复项II问题描述：给你一个有序数组 nums ，请你** 原地** 删除重复出现的元素，使得出现次数超过两次的元素只出现两次 ，返回删除后数组的新长度。不要使用额外的数组空间，你必须在 原地 修改输入数组 并在使用 O(1) 额外空间的条件下完成。解决方案：12345678910111213141516171819class Solution {public: int removeDuplicates(vector<int>& nums) { int n = nums.size(); if (n <= 2) { return n; } int slow = 2, fast = 2; while (fast < n) { if (nums[slow - 2] != nums[fast]) { nums[slow] = nums[fa ...</div></div></div><div class="recent-post-item"><div class="recent-post-info no-cover"><a class="article-title" href="/2023/09/05/%E2%80%9C%E6%88%91%E7%9A%84%E7%AC%AC%E4%BA%94%E7%AF%87%E5%8D%9A%E5%AE%A2%E2%80%9D/" title="“我的第五篇博客”">“我的第五篇博客”</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">Created</span><time datetime="2023-09-05T01:37:41.000Z" title="Created 2023-09-05 09:37:41">2023-09-05</time></span></div><div class="content">删除排序数组中的重复项问题描述：给你一个升序排列 的数组nums ，请你原地删除重复出现的元素，使每个元素只出现一次 ，返回删除后数组的新长度。元素的相对顺序应该保持一致 。然后返回nums中唯一元素的个数。解决思路：依然采用双指针的做法，利用数组有序的特点，可以通过双指针的方法删除重复元素。如果数组 nums 的长度为0，则数组不包含任何元素，因此返回0。当数组 nums 的长度大于 0 时，数组中至少包含一个元素，在删除重复元素之后也至少剩下一个元素，因此 nums[0] 保持原状即可，从下标 1 开始删除重复元素。123456789101112131415161718class Solution {public: int removeDuplicates(vector<int>& nums) { int n = nums.size(); if (n == 0) { return 0; } int fast = 1, slow = 1; ...</div></div></div><div class="recent-post-item"><div class="recent-post-info no-cover"><a class="article-title" href="/2023/09/04/%E6%88%91%E7%9A%84%E7%AC%AC%E5%9B%9B%E7%AF%87%E5%8D%9A%E5%AE%A2/" title="我的第四篇博客">我的第四篇博客</a><div class="article-meta-wrap"><span class="post-meta-date"><i class="far fa-calendar-alt"></i><span class="article-meta-label">Created</span><time datetime="2023-09-04T05:43:38.000Z" title="Created 2023-09-04 13:43:38">2023-09-04</time></span></div><div class="content">移除元素问题描述：给你一个数组nums和一个值val,你需要原地移除所有数值等于val的元素，并返回移除后数组的新长度，不要使用额外的数组空间。解决思路：比较容易想到的直观的思路直接在旧数组上面修改，将输出的数组直接写在输入数组上可以使用双指针1234567891011121314class Solution {public: int removeElement(vector<int>& nums, int val) { int n = nums.size(); int left = 0; for (int right = 0; right < n; right++) { if (nums[right] != val) { nums[left] = nums[right]; left++; } } return left; ...</div></div></div><nav id="pagination"><div class="pagination"><span class="page-number current">1</span><a class="page-number" href="/page/2/#content-inner">2</a><a class="extend next" rel="next" href="/page/2/#content-inner"><i class="fas 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