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<p>ARTS打卡第二周</p>
<p>A: leetcode 3Sum， Midum</p>
<p>R：http断点续传协议TUS</p>
<p>T: 利用git commit 回滚代码，阅读庞大项目</p>
<p>S：go-nats代码阅读</p>
<a id="more"></a>
<p>[TOC]</p>
<h3 id="A-leetcode-3Sum，-Midum">
<a href="#A-leetcode-3Sum，-Midum" class="headerlink" title="A: leetcode 3Sum， Midum"></a>A: leetcode 3Sum， Midum</h3>
<p>实际做了3道题，不过
<code>intToRoman</code>和
<code>romanToInt</code>这两题难度较低，意义不大，于是尝试这道
<code>3sum</code>
</p>
<p>自己的实现：</p>
<details>
<br>
<summary>自己的实现</summary>
<br>
<br>
<figure class="highlight go">
<table>
<tr>
<td class="gutter">
<pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br></pre>
</td>
<td class="code">
<pre><span class="line"> </span><br><span class="line"><span class="function"><span class="keyword">func</span> <span class="title">threeSum</span><span class="params">(nums []<span class="keyword">int</span>)</span> [][]<span class="title">int</span></span> {</span><br><span class="line"> <span class="keyword">if</span> <span class="built_in">len</span>(nums) == <span class="number">0</span> {</span><br><span class="line"> <span class="keyword">return</span> [][]<span class="keyword">int</span>{}</span><br><span class="line"> }</span><br><span class="line"> sort.Slice(nums, <span class="function"><span class="keyword">func</span><span class="params">(i, j <span class="keyword">int</span>)</span> <span class="title">bool</span></span> {</span><br><span class="line"> <span class="keyword">return</span> nums[j] > nums[i]</span><br><span class="line"> })</span><br><span class="line"> positiveMap := <span class="keyword">map</span>[<span class="keyword">int</span>]<span class="keyword">int</span>{}</span><br><span class="line"> <span class="keyword">for</span> index, i := <span class="keyword">range</span> nums {</span><br><span class="line"> <span class="keyword">if</span> i >= <span class="number">0</span> {</span><br><span class="line"> positiveMap[i] = index</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> result := [][]<span class="keyword">int</span>{}</span><br><span class="line"> currA := nums[<span class="number">0</span>] - <span class="number">1</span></span><br><span class="line"> <span class="keyword">for</span> i := <span class="number">0</span>; i < <span class="built_in">len</span>(nums)<span class="number">-1</span>; i++ {</span><br><span class="line"> <span class="keyword">if</span> nums[i] > <span class="number">0</span> {</span><br><span class="line"> <span class="keyword">break</span></span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">if</span> currA == nums[i] {</span><br><span class="line"> <span class="keyword">continue</span></span><br><span class="line"> }</span><br><span class="line"> currA = nums[i]</span><br><span class="line"> currB := nums[<span class="number">0</span>] - <span class="number">1</span></span><br><span class="line"> <span class="keyword">for</span> j := i + <span class="number">1</span>; j < <span class="built_in">len</span>(nums); j++ {</span><br><span class="line"> <span class="keyword">if</span> currB == nums[j] {</span><br><span class="line"> <span class="keyword">continue</span></span><br><span class="line"> }</span><br><span class="line"> currB = nums[j]</span><br><span class="line"> target := <span class="number">0</span> - (nums[i] + nums[j])</span><br><span class="line"> <span class="keyword">if</span> c, ok := positiveMap[target]; ok {</span><br><span class="line"> <span class="keyword">if</span> c > j {</span><br><span class="line"> result = <span class="built_in">append</span>(result, []<span class="keyword">int</span>{</span><br><span class="line"> nums[i], nums[j], target,</span><br><span class="line"> })</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> result</span><br><span class="line">}</span><br></pre>
</td>
</tr>
</table>
</figure>
<br>
<br>
</details>
<p> 运行结果：</p>
<blockquote>
<p>Runtime: 936 ms, faster than 48.61% of Go online submissions
for 3Sum.</p>
<p>Memory Usage: 340.6 MB, less than 30.93% of Go online submissions
for 3Sum.</p>
</blockquote>
<p>参考大神代码后的结果：</p>
<figure class="highlight go">
<table>
<tr>
<td class="gutter">
<pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br></pre>
</td>
<td class="code">
<pre><span class="line"><span class="function"><span class="keyword">func</span> <span class="title">threeSum</span><span class="params">(nums []<span class="keyword">int</span>)</span> [][]<span class="title">int</span></span> {</span><br><span class="line"> result := [][]<span class="keyword">int</span>{}</span><br><span class="line"> sort.Slice(nums, <span class="function"><span class="keyword">func</span><span class="params">(i, j <span class="keyword">int</span>)</span> <span class="title">bool</span></span> {</span><br><span class="line"> <span class="keyword">return</span> nums[i] < nums[j]</span><br><span class="line"> })</span><br><span class="line"> <span class="keyword">for</span> i := <span class="number">0</span>; i < <span class="built_in">len</span>(nums)<span class="number">-2</span> && nums[i] <= <span class="number">0</span>; i++ {</span><br><span class="line"> <span class="keyword">if</span> i > <span class="number">0</span> && nums[i] == nums[i<span class="number">-1</span>] { <span class="comment">// 重复的跳过</span></span><br><span class="line"> <span class="keyword">continue</span></span><br><span class="line"> }</span><br><span class="line"> <span class="comment">// s指向数组开头即最小值，e指向最后，即最大值</span></span><br><span class="line"> s, e, target := i+<span class="number">1</span>, <span class="built_in">len</span>(nums)<span class="number">-1</span>, <span class="number">0</span>-nums[i]</span><br><span class="line"> <span class="keyword">for</span> s < e {</span><br><span class="line"> <span class="comment">// fmt.Printf("nums %+v\n s: %d, e:%d\n", nums, s, e)</span></span><br><span class="line"> <span class="keyword">if</span> nums[s]+nums[e] == target {</span><br><span class="line"> result = <span class="built_in">append</span>(result, []<span class="keyword">int</span>{</span><br><span class="line"> nums[i], nums[s], nums[e],</span><br><span class="line"> })</span><br><span class="line"> <span class="keyword">for</span> s < e && nums[s] == nums[s+<span class="number">1</span>] {</span><br><span class="line"> s++</span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">for</span> s < e && nums[e] == nums[e<span class="number">-1</span>] {</span><br><span class="line"> e--</span><br><span class="line"> }</span><br><span class="line"> s++</span><br><span class="line"> e--</span><br><span class="line"> } <span class="keyword">else</span> <span class="keyword">if</span> nums[s]+nums[e] > target { <span class="comment">// 结果大于逾期，需要最大值减小</span></span><br><span class="line"> e--</span><br><span class="line"> } <span class="keyword">else</span> {</span><br><span class="line"> s++</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"></span><br><span class="line"> }</span><br><span class="line"> <span class="keyword">return</span> result</span><br><span class="line">}</span><br></pre>
</td>
</tr>
</table>
</figure>
<p>运行结果：</p>
<blockquote>
<p>Runtime: 868 ms, faster than 91.25% of Go online submissions
for 3Sum.</p>
<p>Memory Usage: 205.1 MB, less than 98.97% of Go online submissions
for 3Sum.</p>
</blockquote>
<h3 id="R：http断点续传协议TUS">
<a href="#R：http断点续传协议TUS" class="headerlink" title="R：http断点续传协议TUS"></a>R：http断点续传协议TUS</h3>
<p>
<a href="https://golangbot.com/understanding-tus/" target="_blank" rel="external nofollow noopener noreferrer">Resumable file uploader: Understanding tus protocol</a>
</p>
<p>通过http方式解决大文件断点续传问题，主要通过三个步骤实现：</p>
<ul>
<li>http
<code>POST</code>方法,创建文件，返回rest格式的文件地址</li>
<li>http
<code>PATCH</code>方案更新文件，每次传入更新的
<code>Upload-Offset</code>和
<code>Content-Length</code>,</li>
<li>http
<code>HEAD</code>返回文件更新位置，即最后一次
<code>PATCH</code>的位置</li>
</ul>
<h3 id="T-利用git-commit-回滚代码，阅读庞大项目">
<a href="#T-利用git-commit-回滚代码，阅读庞大项目" class="headerlink" title="T: 利用git commit 回滚代码，阅读庞大项目"></a>T: 利用git commit 回滚代码，阅读庞大项目</h3>
<p>这个技巧来自于公众号一个大神的推荐，
<a href="https://mp.weixin.qq.com/s/LsGEr_RJYfAbGBBlRxJH-Q" target="_blank" rel="external nofollow noopener noreferrer">开源代码学习技巧</a>
</p>
<p>内容主要就是clone整个项目后，可以通过提交历史找到一个较早的可用提交，个人经验可以看tag名称，比如v0.01，v0.24,这种，或是看具体的commit
message，这样可以看到最小功能的项目代码，理解最初意图，主要是降低阅读代价，然后不断向后移动提交位置，不仅可以逐步阅读完代码，还可以看到作者在每次更新的主要内容。</p>
<h3 id="S：go-nats代码阅读">
<a href="#S：go-nats代码阅读" class="headerlink" title="S：go-nats代码阅读"></a>S：go-nats代码阅读</h3>
<p>用上面提到的技巧，自己整理了一下第一版go-nats代码，很快就看懂了，接下来会阅读
<code>v0.6</code>，这样逐步完成整个项目阅读</p>
<p>
<img src="image-20190414102359041.png" alt="image-20190414102359041">
</p>
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<a class="nav-link" href="#A-leetcode-3Sum，-Midum">
<span class="nav-number">1.</span>
<span class="nav-text">
A: leetcode 3Sum， Midum</span>
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<li class="nav-item nav-level-3">
<a class="nav-link" href="#R：http断点续传协议TUS">
<span class="nav-number">2.</span>
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R：http断点续传协议TUS</span>
</a>
</li>
<li class="nav-item nav-level-3">
<a class="nav-link" href="#T-利用git-commit-回滚代码，阅读庞大项目">
<span class="nav-number">3.</span>
<span class="nav-text">
T: 利用git commit 回滚代码，阅读庞大项目</span>
</a>
</li>
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<a class="nav-link" href="#S：go-nats代码阅读">
<span class="nav-number">4.</span>
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S：go-nats代码阅读</span>
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