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32.常考算法题解析.htm
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<!DOCTYPE html>
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<html lang="zh-CN"><head><meta http-equiv="Content-Type" content="text/html; charset=UTF-8"><meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1"><meta name="viewport" content="width=device-width,initial-scale=1,user-scalable=no,viewport-fit=cover"><meta name="google-site-verification" content="cCHsgG9ktuCTgWgYfqCJql8AeR4gAne4DTZqztPoirE"><meta name="apple-itunes-app" content="app-id=987739104"><meta name="baidu-site-verification" content="qiK2a1kcFc"><meta name="360-site-verification" content="4c3c7d57d59f0e1a308462fbc7fd7e51"><meta name="sogou_site_verification" content="c49WUDZczQ"><style>body {
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data-v-1596580a=""><div class="btn-ok" data-v-1596580a="">知道了</div><div class="btn-ok btn-link" data-v-1596580a="">前往小册首页</div></div></div><div class="model fail" style="display:none;" data-v-1596580a=""><div class="heading" data-v-1596580a="">领取失败</div><div class="content-text" data-v-1596580a="">本活动仅适用于小册新用户</div><div class="btn-ok" data-v-1596580a="">知道了</div></div></div><!----><!----><div class="bind-phone-number-modal-box" data-v-b421b7f4="" data-v-35e99930=""><div st:block="bindPhoneNumberModal" class="modal-mask" style="display:none;" data-v-b421b7f4=""></div><form class="bind-phone-number-form" style="display:none;" data-v-b421b7f4=""><i title="关闭" class="close-btn ion-close-round" data-v-b421b7f4=""></i><h1 class="title" data-v-b421b7f4="">提示</h1><h2 class="hint" data-v-b421b7f4="">根据我国<a href="http://www.npc.gov.cn/npc/xinwen/2016-11/07/content_2001605.htm" target="_blank" style="color:#007fff">《网络安全法》</a>，您需要绑定手机号后才可在掘金社区内发布内容。</h2><div class="input-group" data-v-b421b7f4=""><div class="input-box" data-v-b421b7f4=""><input maxlength="64" placeholder="请输入要绑定的手机号码" value="" class="input" data-v-b421b7f4=""></div><!----><!----></div><button st:name="bindBtn" class="btn" data-v-b421b7f4="">绑定手机</button></form></div></div><!----><div data-v-54b8e7d6="" data-v-35e99930="" class="book-read-view"><section data-v-54b8e7d6="" class="book-section"><div data-v-54b8e7d6="" class="book-summary"><div data-v-54b8e7d6="" class="book-summary-masker"></div><div data-v-54b8e7d6="" class="book-summary-inner"><div data-v-54b8e7d6="" class="book-summary__header"><a data-v-54b8e7d6="" href="https://juejin.im/books" target="" rel="" class="logo"><img data-v-54b8e7d6="" src="./32.常考算法题解析_files/logo.a7995ad.svg"></a><div data-v-54b8e7d6="" class="label">小册</div><!----></div><!----><div data-v-d0eb2184="" data-v-54b8e7d6="" class="book-directory book-directory bought"><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">1</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">小册食用指南</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">2</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">JS 基础知识点及常考面试题（一）</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">3</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">JS 基础知识点及常考面试题（二）</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">4</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">ES6 知识点及常考面试题</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">5</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">JS 异步编程及常考面试题</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">6</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">手写 Promise</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">7</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">Event Loop</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">8</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">JS 进阶知识点及常考面试题</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">9</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">JS 思考题</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">10</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">DevTools Tips</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">11</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">浏览器基础知识点及常考面试题</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">12</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">浏览器缓存机制</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">13</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">浏览器渲染原理</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">14</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">安全防范知识点</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">15</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">从 V8 中看 JS 性能优化</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">16</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">性能优化琐碎事</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">17</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">Webpack 性能优化</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">18</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">实现小型打包工具</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">19</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">React 和 Vue 两大框架之间的相爱相杀</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">20</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">Vue 常考基础知识点</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">21</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">Vue 常考进阶知识点</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">22</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">React 常考基础知识点</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">23</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">React 常考进阶知识点</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">24</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">监控</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">25</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">UDP</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">26</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">TCP</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">27</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">HTTP 及 TLS</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">28</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">HTTP/2 及 HTTP/3</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">29</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">输入 URL 到页面渲染的整个流程</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">30</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">设计模式</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link read"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">31</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">常见数据结构</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section route-active section-link"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">32</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">常考算法题解析</div><!----><!----></div><!----></a><a data-v-d0eb2184="" class="section section-link"><div data-v-d0eb2184="" class="step"><div data-v-d0eb2184="" class="step-btn">33</div></div><div data-v-d0eb2184="" class="center"><div data-v-d0eb2184="" class="title">CSS 常考面试题资料</div><!----><!----></div><!----></a><a 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<p>这一章节依托于上一章节的内容，毕竟了解了数据结构我们才能写出更好的算法。</p>
<p>对于大部分公司的面试来说，排序的内容已经足以应付了，由此为了更好的符合大众需求，排序的内容是最多的。当然如果你还想冲击更好的公司，那么整一个章节的内容都是需要掌握的。对于字节跳动这类十分看重算法的公司来说，这一章节是远远不够的，<a target="_blank" href="https://link.juejin.im/?target=https%3A%2F%2Fbook.douban.com%2Fsubject%2F6966465%2F" rel="nofollow noopener noreferrer">剑指Offer</a>应该是你更好的选择。</p>
<blockquote class="warning"><p>这一章节的内容信息量会很大，不适合在非电脑环境下阅读，请各位打开代码编辑器，一行行的敲代码，单纯阅读是学习不了算法的。
</p></blockquote><p>另外学习算法的时候，有一个可视化界面会相对减少点学习的难度，具体可以阅读 <a target="_blank" href="https://link.juejin.im/?target=https%3A%2F%2Fgithub.com%2Falgorithm-visualizer%2Falgorithm-visualizer" rel="nofollow noopener noreferrer">algorithm-visualizer</a> 这个仓库。</p>
<h2 class="heading" data-id="heading-1">位运算</h2>
<p>在进入正题之前，我们先来学习一下位运算的内容。因为位运算在算法中很有用，速度可以比四则运算快很多。</p>
<p>在学习位运算之前应该知道十进制如何转二进制，二进制如何转十进制。这里说明下简单的计算方式</p>
<ul>
<li>十进制 <code>33</code> 可以看成是 <code>32 + 1</code> ，并且 <code>33</code> 应该是六位二进制的（因为 <code>33</code> 近似 <code>32</code>，而 <code>32</code> 是 2 的五次方，所以是六位），那么 十进制 <code>33</code> 就是 <code>100001</code> ，只要是 2 的次方，那么就是 1否则都为 0</li>
<li>那么二进制 <code>100001</code> 同理，首位是 <code>2^5</code> ，末位是 <code>2^0</code> ，相加得出 33</li>
</ul>
<h3 class="heading" data-id="heading-2">左移 <<</h3>
<pre><code class="hljs js" lang="js"><span class="hljs-number">10</span> << <span class="hljs-number">1</span> <span class="hljs-comment">// -> 20</span>
</code></pre><p>左移就是将二进制全部往左移动，<code>10</code> 在二进制中表示为 <code>1010</code> ，左移一位后变成 <code>10100</code> ，转换为十进制也就是 20，所以基本可以把左移看成以下公式 <code>a * (2 ^ b)</code></p>
<h3 class="heading" data-id="heading-3">算数右移 >></h3>
<pre><code class="hljs js" lang="js"><span class="hljs-number">10</span> >> <span class="hljs-number">1</span> <span class="hljs-comment">// -> 5</span>
</code></pre><p>算数右移就是将二进制全部往右移动并去除多余的右边，<code>10</code> 在二进制中表示为 <code>1010</code> ，右移一位后变成 <code>101</code> ，转换为十进制也就是 5，所以基本可以把右移看成以下公式 <code>int v = a / (2 ^ b)</code></p>
<p>右移很好用，比如可以用在二分算法中取中间值</p>
<pre><code class="hljs js" lang="js"><span class="hljs-number">13</span> >> <span class="hljs-number">1</span> <span class="hljs-comment">// -> 6</span>
</code></pre><h3 class="heading" data-id="heading-4">按位操作</h3>
<p><strong>按位与</strong></p>
<p>每一位都为 1，结果才为 1</p>
<pre><code class="hljs js" lang="js"><span class="hljs-number">8</span> & <span class="hljs-number">7</span> <span class="hljs-comment">// -> 0</span>
<span class="hljs-comment">// 1000 & 0111 -> 0000 -> 0</span>
</code></pre><p><strong>按位或</strong></p>
<p>其中一位为 1，结果就是 1</p>
<pre><code class="hljs js" lang="js"><span class="hljs-number">8</span> | <span class="hljs-number">7</span> <span class="hljs-comment">// -> 15</span>
<span class="hljs-comment">// 1000 | 0111 -> 1111 -> 15</span>
</code></pre><p><strong>按位异或</strong></p>
<p>每一位都不同，结果才为 1</p>
<pre><code class="hljs js" lang="js"><span class="hljs-number">8</span> ^ <span class="hljs-number">7</span> <span class="hljs-comment">// -> 15</span>
<span class="hljs-number">8</span> ^ <span class="hljs-number">8</span> <span class="hljs-comment">// -> 0</span>
<span class="hljs-comment">// 1000 ^ 0111 -> 1111 -> 15</span>
<span class="hljs-comment">// 1000 ^ 1000 -> 0000 -> 0</span>
</code></pre><p>从以上代码中可以发现按位异或就是不进位加法</p>
<p><strong>面试题</strong>：两个数不使用四则运算得出和</p>
<p>这道题中可以按位异或，因为按位异或就是不进位加法，<code>8 ^ 8 = 0</code> 如果进位了，就是 16 了，所以我们只需要将两个数进行异或操作，然后进位。那么也就是说两个二进制都是 1 的位置，左边应该有一个进位 1，所以可以得出以下公式 <code>a + b = (a ^ b) + ((a & b) << 1)</code> ，然后通过迭代的方式模拟加法</p>
<pre><code class="hljs js" lang="js"><span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">sum</span>(<span class="hljs-params">a, b</span>) </span>{
<span class="hljs-keyword">if</span> (a == <span class="hljs-number">0</span>) <span class="hljs-keyword">return</span> b
<span class="hljs-keyword">if</span> (b == <span class="hljs-number">0</span>) <span class="hljs-keyword">return</span> a
<span class="hljs-keyword">let</span> newA = a ^ b
<span class="hljs-keyword">let</span> newB = (a & b) << <span class="hljs-number">1</span>
<span class="hljs-keyword">return</span> sum(newA, newB)
}
</code></pre><h2 class="heading" data-id="heading-5">排序</h2>
<p>以下两个函数是排序中会用到的通用函数，就不一一写了</p>
<pre><code class="hljs js" lang="js"><span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">checkArray</span>(<span class="hljs-params">array</span>) </span>{
<span class="hljs-keyword">return</span> <span class="hljs-built_in">Array</span>.isArray(array)
}
<span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">swap</span>(<span class="hljs-params">array, left, right</span>) </span>{
<span class="hljs-keyword">let</span> rightValue = array[right]
array[right] = array[left]
array[left] = rightValue
}
</code></pre><h3 class="heading" data-id="heading-6">冒泡排序</h3>
<p>冒泡排序的原理如下，从第一个元素开始，把当前元素和下一个索引元素进行比较。如果当前元素大，那么就交换位置，重复操作直到比较到最后一个元素，那么此时最后一个元素就是该数组中最大的数。下一轮重复以上操作，但是此时最后一个元素已经是最大数了，所以不需要再比较最后一个元素，只需要比较到 <code>length - 2</code> 的位置。</p>
<div align="center">
<img width="500" class="lazyload inited" data-src="https://user-gold-cdn.xitu.io/2018/4/12/162b895b452b306c?imageslim" data-width="670" data-height="508" src="data:image/svg+xml;utf8,<?xml version="1.0"?><svg xmlns="http://www.w3.org/2000/svg" version="1.1" width="670" height="508"></svg>">
</div>
<p>以下是实现该算法的代码</p>
<pre><code class="hljs js" lang="js"><span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">bubble</span>(<span class="hljs-params">array</span>) </span>{
checkArray(array);
<span class="hljs-keyword">for</span> (<span class="hljs-keyword">let</span> i = array.length - <span class="hljs-number">1</span>; i > <span class="hljs-number">0</span>; i--) {
<span class="hljs-comment">// 从 0 到 `length - 1` 遍历</span>
<span class="hljs-keyword">for</span> (<span class="hljs-keyword">let</span> j = <span class="hljs-number">0</span>; j < i; j++) {
<span class="hljs-keyword">if</span> (array[j] > array[j + <span class="hljs-number">1</span>]) swap(array, j, j + <span class="hljs-number">1</span>)
}
}
<span class="hljs-keyword">return</span> array;
}
</code></pre><p>该算法的操作次数是一个等差数列 <code>n + (n - 1) + (n - 2) + 1</code> ，去掉常数项以后得出时间复杂度是 O(n * n)</p>
<h3 class="heading" data-id="heading-7">插入排序</h3>
<p>插入排序的原理如下。第一个元素默认是已排序元素，取出下一个元素和当前元素比较，如果当前元素大就交换位置。那么此时第一个元素就是当前的最小数，所以下次取出操作从第三个元素开始，向前对比，重复之前的操作。</p>
<div align="center"><img style="display:block;margin: 0 auto" width="500" class="lazyload inited" data-src="https://user-gold-cdn.xitu.io/2018/4/12/162b895c7e59dcd1?imageslim" data-width="670" data-height="508" src="data:image/svg+xml;utf8,<?xml version="1.0"?><svg xmlns="http://www.w3.org/2000/svg" version="1.1" width="670" height="508"></svg>"></div>
<p>以下是实现该算法的代码</p>
<pre><code class="hljs js" lang="js"><span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">insertion</span>(<span class="hljs-params">array</span>) </span>{
<span class="hljs-keyword">if</span> (!checkArray(array)) <span class="hljs-keyword">return</span>
<span class="hljs-keyword">for</span> (<span class="hljs-keyword">let</span> i = <span class="hljs-number">1</span>; i < array.length; i++) {
<span class="hljs-keyword">for</span> (<span class="hljs-keyword">let</span> j = i - <span class="hljs-number">1</span>; j >= <span class="hljs-number">0</span> && array[j] > array[j + <span class="hljs-number">1</span>]; j--)
swap(array, j, j + <span class="hljs-number">1</span>);
}
<span class="hljs-keyword">return</span> array;
}
</code></pre><p>该算法的操作次数是一个等差数列 <code>n + (n - 1) + (n - 2) + 1</code> ，去掉常数项以后得出时间复杂度是 O(n * n)</p>
<h3 class="heading" data-id="heading-8">选择排序</h3>
<p>选择排序的原理如下。遍历数组，设置最小值的索引为 0，如果取出的值比当前最小值小，就替换最小值索引，遍历完成后，将第一个元素和最小值索引上的值交换。如上操作后，第一个元素就是数组中的最小值，下次遍历就可以从索引 1 开始重复上述操作。</p>
<div align="center"><img style="display:block;margin: 0 auto" width="500" class="lazyload inited" data-src="https://user-gold-cdn.xitu.io/2018/4/13/162bc8ea14567e2e?imageslim" data-width="670" data-height="508" src="data:image/svg+xml;utf8,<?xml version="1.0"?><svg xmlns="http://www.w3.org/2000/svg" version="1.1" width="670" height="508"></svg>"></div>
<p>以下是实现该算法的代码</p>
<pre><code class="hljs js" lang="js"><span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">selection</span>(<span class="hljs-params">array</span>) </span>{
<span class="hljs-keyword">if</span> (!checkArray(array)) <span class="hljs-keyword">return</span>
<span class="hljs-keyword">for</span> (<span class="hljs-keyword">let</span> i = <span class="hljs-number">0</span>; i < array.length - <span class="hljs-number">1</span>; i++) {
<span class="hljs-keyword">let</span> minIndex = i;
<span class="hljs-keyword">for</span> (<span class="hljs-keyword">let</span> j = i + <span class="hljs-number">1</span>; j < array.length; j++) {
minIndex = array[j] < array[minIndex] ? j : minIndex;
}
swap(array, i, minIndex);
}
<span class="hljs-keyword">return</span> array;
}
</code></pre><p>该算法的操作次数是一个等差数列 <code>n + (n - 1) + (n - 2) + 1</code> ，去掉常数项以后得出时间复杂度是 O(n * n)</p>
<h3 class="heading" data-id="heading-9">归并排序</h3>
<p>归并排序的原理如下。递归的将数组两两分开直到最多包含两个元素，然后将数组排序合并，最终合并为排序好的数组。假设我有一组数组 <code>[3, 1, 2, 8, 9, 7, 6]</code>，中间数索引是 3，先排序数组 <code>[3, 1, 2, 8]</code> 。在这个左边数组上，继续拆分直到变成数组包含两个元素（如果数组长度是奇数的话，会有一个拆分数组只包含一个元素）。然后排序数组 <code>[3, 1]</code> 和 <code>[2, 8]</code> ，然后再排序数组 <code>[1, 3, 2, 8]</code> ，这样左边数组就排序完成，然后按照以上思路排序右边数组，最后将数组 <code>[1, 2, 3, 8]</code> 和 <code>[6, 7, 9]</code> 排序。</p>
<div align="center"><img width="500" class="lazyload inited" data-src="https://user-gold-cdn.xitu.io/2018/4/13/162be13c7e30bd86?imageslim" data-width="896" data-height="1008" src="data:image/svg+xml;utf8,<?xml version="1.0"?><svg xmlns="http://www.w3.org/2000/svg" version="1.1" width="896" height="1008"></svg>"></div>
<p>以下是实现该算法的代码</p>
<pre><code class="hljs js" lang="js"><span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">sort</span>(<span class="hljs-params">array</span>) </span>{
<span class="hljs-keyword">if</span> (!checkArray(array)) <span class="hljs-keyword">return</span>
mergeSort(array, <span class="hljs-number">0</span>, array.length - <span class="hljs-number">1</span>);
<span class="hljs-keyword">return</span> array;
}
<span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">mergeSort</span>(<span class="hljs-params">array, left, right</span>) </span>{
<span class="hljs-comment">// 左右索引相同说明已经只有一个数</span>
<span class="hljs-keyword">if</span> (left === right) <span class="hljs-keyword">return</span>;
<span class="hljs-comment">// 等同于 `left + (right - left) / 2`</span>
<span class="hljs-comment">// 相比 `(left + right) / 2` 来说更加安全，不会溢出</span>
<span class="hljs-comment">// 使用位运算是因为位运算比四则运算快</span>
<span class="hljs-keyword">let</span> mid = <span class="hljs-built_in">parseInt</span>(left + ((right - left) >> <span class="hljs-number">1</span>));
mergeSort(array, left, mid);
mergeSort(array, mid + <span class="hljs-number">1</span>, right);
<span class="hljs-keyword">let</span> help = [];
<span class="hljs-keyword">let</span> i = <span class="hljs-number">0</span>;
<span class="hljs-keyword">let</span> p1 = left;
<span class="hljs-keyword">let</span> p2 = mid + <span class="hljs-number">1</span>;
<span class="hljs-keyword">while</span> (p1 <= mid && p2 <= right) {
help[i++] = array[p1] < array[p2] ? array[p1++] : array[p2++];
}
<span class="hljs-keyword">while</span> (p1 <= mid) {
help[i++] = array[p1++];
}
<span class="hljs-keyword">while</span> (p2 <= right) {
help[i++] = array[p2++];
}
<span class="hljs-keyword">for</span> (<span class="hljs-keyword">let</span> i = <span class="hljs-number">0</span>; i < help.length; i++) {
array[left + i] = help[i];
}
<span class="hljs-keyword">return</span> array;
}
</code></pre><p>以上算法使用了递归的思想。递归的本质就是压栈，每递归执行一次函数，就将该函数的信息（比如参数，内部的变量，执行到的行数）压栈，直到遇到终止条件，然后出栈并继续执行函数。对于以上递归函数的调用轨迹如下</p>
<pre><code class="hljs js" lang="js">mergeSort(data, <span class="hljs-number">0</span>, <span class="hljs-number">6</span>) <span class="hljs-comment">// mid = 3</span>
mergeSort(data, <span class="hljs-number">0</span>, <span class="hljs-number">3</span>) <span class="hljs-comment">// mid = 1</span>
mergeSort(data, <span class="hljs-number">0</span>, <span class="hljs-number">1</span>) <span class="hljs-comment">// mid = 0</span>
mergeSort(data, <span class="hljs-number">0</span>, <span class="hljs-number">0</span>) <span class="hljs-comment">// 遇到终止，回退到上一步</span>
mergeSort(data, <span class="hljs-number">1</span>, <span class="hljs-number">1</span>) <span class="hljs-comment">// 遇到终止，回退到上一步</span>
<span class="hljs-comment">// 排序 p1 = 0, p2 = mid + 1 = 1</span>
<span class="hljs-comment">// 回退到 `mergeSort(data, 0, 3)` 执行下一个递归</span>
mergeSort(<span class="hljs-number">2</span>, <span class="hljs-number">3</span>) <span class="hljs-comment">// mid = 2</span>
mergeSort(<span class="hljs-number">3</span>, <span class="hljs-number">3</span>) <span class="hljs-comment">// 遇到终止，回退到上一步</span>
<span class="hljs-comment">// 排序 p1 = 2, p2 = mid + 1 = 3</span>
<span class="hljs-comment">// 回退到 `mergeSort(data, 0, 3)` 执行合并逻辑</span>
<span class="hljs-comment">// 排序 p1 = 0, p2 = mid + 1 = 2</span>
<span class="hljs-comment">// 执行完毕回退</span>
<span class="hljs-comment">// 左边数组排序完毕，右边也是如上轨迹</span>
</code></pre><p>该算法的操作次数是可以这样计算：递归了两次，每次数据量是数组的一半，并且最后把整个数组迭代了一次，所以得出表达式 <code>2T(N / 2) + T(N)</code> （T 代表时间，N 代表数据量）。根据该表达式可以套用 <a target="_blank" href="https://link.juejin.im/?target=https%3A%2F%2Fwww.wikiwand.com%2Fzh-hans%2F%25E4%25B8%25BB%25E5%25AE%259A%25E7%2590%2586" rel="nofollow noopener noreferrer">该公式</a> 得出时间复杂度为 <code>O(N * logN)</code></p>
<h3 class="heading" data-id="heading-10">快排</h3>
<p>快排的原理如下。随机选取一个数组中的值作为基准值，从左至右取值与基准值对比大小。比基准值小的放数组左边，大的放右边，对比完成后将基准值和第一个比基准值大的值交换位置。然后将数组以基准值的位置分为两部分，继续递归以上操作。</p>
<div align="center"><img width="500" class="lazyload inited" data-src="https://user-gold-cdn.xitu.io/2018/4/16/162cd23e69ca9ea3?imageslim" data-width="824" data-height="506" src="data:image/svg+xml;utf8,<?xml version="1.0"?><svg xmlns="http://www.w3.org/2000/svg" version="1.1" width="824" height="506"></svg>"></div>
<p>以下是实现该算法的代码</p>
<pre><code class="hljs js" lang="js"><span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">sort</span>(<span class="hljs-params">array</span>) </span>{
<span class="hljs-keyword">if</span> (!checkArray(array)) <span class="hljs-keyword">return</span>
quickSort(array, <span class="hljs-number">0</span>, array.length - <span class="hljs-number">1</span>);
<span class="hljs-keyword">return</span> array;
}
<span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">quickSort</span>(<span class="hljs-params">array, left, right</span>) </span>{
<span class="hljs-keyword">if</span> (left < right) {
swap(array, , right)
<span class="hljs-comment">// 随机取值，然后和末尾交换，这样做比固定取一个位置的复杂度略低</span>
<span class="hljs-keyword">let</span> indexs = part(array, <span class="hljs-built_in">parseInt</span>(<span class="hljs-built_in">Math</span>.random() * (right - left + <span class="hljs-number">1</span>)) + left, right);
quickSort(array, left, indexs[<span class="hljs-number">0</span>]);
quickSort(array, indexs[<span class="hljs-number">1</span>] + <span class="hljs-number">1</span>, right);
}
}
<span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">part</span>(<span class="hljs-params">array, left, right</span>) </span>{
<span class="hljs-keyword">let</span> less = left - <span class="hljs-number">1</span>;
<span class="hljs-keyword">let</span> more = right;
<span class="hljs-keyword">while</span> (left < more) {
<span class="hljs-keyword">if</span> (array[left] < array[right]) {
<span class="hljs-comment">// 当前值比基准值小，`less` 和 `left` 都加一</span>
++less;
++left;
} <span class="hljs-keyword">else</span> <span class="hljs-keyword">if</span> (array[left] > array[right]) {
<span class="hljs-comment">// 当前值比基准值大，将当前值和右边的值交换</span>
<span class="hljs-comment">// 并且不改变 `left`，因为当前换过来的值还没有判断过大小</span>
swap(array, --more, left);
} <span class="hljs-keyword">else</span> {
<span class="hljs-comment">// 和基准值相同，只移动下标</span>
left++;
}
}
<span class="hljs-comment">// 将基准值和比基准值大的第一个值交换位置</span>
<span class="hljs-comment">// 这样数组就变成 `[比基准值小, 基准值, 比基准值大]`</span>
swap(array, right, more);
<span class="hljs-keyword">return</span> [less, more];
}
</code></pre><p>该算法的复杂度和归并排序是相同的，但是额外空间复杂度比归并排序少，只需 O(logN)，并且相比归并排序来说，所需的常数时间也更少。</p>
<h4 class="heading" data-id="heading-11">面试题</h4>
<p><strong>Sort Colors</strong>：该题目来自 <a target="_blank" href="https://link.juejin.im/?target=https%3A%2F%2Fleetcode.com%2Fproblems%2Fsort-colors%2Fdescription%2F" rel="nofollow noopener noreferrer">LeetCode</a>，题目需要我们将 <code>[2,0,2,1,1,0]</code> 排序成 <code>[0,0,1,1,2,2]</code> ，这个问题就可以使用三路快排的思想。</p>
<p>以下是代码实现</p>
<pre><code class="hljs js" lang="js"><span class="hljs-keyword">var</span> sortColors = <span class="hljs-function"><span class="hljs-keyword">function</span>(<span class="hljs-params">nums</span>) </span>{
<span class="hljs-keyword">let</span> left = <span class="hljs-number">-1</span>;
<span class="hljs-keyword">let</span> right = nums.length;
<span class="hljs-keyword">let</span> i = <span class="hljs-number">0</span>;
<span class="hljs-comment">// 下标如果遇到 right，说明已经排序完成</span>
<span class="hljs-keyword">while</span> (i < right) {
<span class="hljs-keyword">if</span> (nums[i] == <span class="hljs-number">0</span>) {
swap(nums, i++, ++left);
} <span class="hljs-keyword">else</span> <span class="hljs-keyword">if</span> (nums[i] == <span class="hljs-number">1</span>) {
i++;
} <span class="hljs-keyword">else</span> {
swap(nums, i, --right);
}
}
};
</code></pre><p><strong>Kth Largest Element in an Array</strong>：该题目来自 <a target="_blank" href="https://link.juejin.im/?target=https%3A%2F%2Fleetcode.com%2Fproblems%2Fkth-largest-element-in-an-array%2Fdescription%2F" rel="nofollow noopener noreferrer">LeetCode</a>，题目需要找出数组中第 K 大的元素，这问题也可以使用快排的思路。并且因为是找出第 K 大元素，所以在分离数组的过程中，可以找出需要的元素在哪边，然后只需要排序相应的一边数组就好。</p>
<p>以下是代码实现</p>
<pre><code class="hljs js" lang="js"><span class="hljs-keyword">var</span> findKthLargest = <span class="hljs-function"><span class="hljs-keyword">function</span>(<span class="hljs-params">nums, k</span>) </span>{
<span class="hljs-keyword">let</span> l = <span class="hljs-number">0</span>
<span class="hljs-keyword">let</span> r = nums.length - <span class="hljs-number">1</span>
<span class="hljs-comment">// 得出第 K 大元素的索引位置</span>
k = nums.length - k
<span class="hljs-keyword">while</span> (l < r) {
<span class="hljs-comment">// 分离数组后获得比基准树大的第一个元素索引</span>
<span class="hljs-keyword">let</span> index = part(nums, l, r)
<span class="hljs-comment">// 判断该索引和 k 的大小</span>
<span class="hljs-keyword">if</span> (index < k) {
l = index + <span class="hljs-number">1</span>
} <span class="hljs-keyword">else</span> <span class="hljs-keyword">if</span> (index > k) {
r = index - <span class="hljs-number">1</span>
} <span class="hljs-keyword">else</span> {
<span class="hljs-keyword">break</span>
}
}
<span class="hljs-keyword">return</span> nums[k]
};
<span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">part</span>(<span class="hljs-params">array, left, right</span>) </span>{
<span class="hljs-keyword">let</span> less = left - <span class="hljs-number">1</span>;
<span class="hljs-keyword">let</span> more = right;
<span class="hljs-keyword">while</span> (left < more) {
<span class="hljs-keyword">if</span> (array[left] < array[right]) {
++less;
++left;
} <span class="hljs-keyword">else</span> <span class="hljs-keyword">if</span> (array[left] > array[right]) {
swap(array, --more, left);
} <span class="hljs-keyword">else</span> {
left++;
}
}
swap(array, right, more);
<span class="hljs-keyword">return</span> more;
}
</code></pre><h3 class="heading" data-id="heading-12">堆排序</h3>
<p>堆排序利用了二叉堆的特性来做，二叉堆通常用数组表示，并且二叉堆是一颗完全二叉树（所有叶节点（最底层的节点）都是从左往右顺序排序，并且其他层的节点都是满的）。二叉堆又分为大根堆与小根堆。</p>
<ul>
<li>大根堆是某个节点的所有子节点的值都比他小</li>
<li>小根堆是某个节点的所有子节点的值都比他大</li>
</ul>
<p>堆排序的原理就是组成一个大根堆或者小根堆。以小根堆为例，某个节点的左边子节点索引是 <code>i * 2 + 1</code>，右边是 <code>i * 2 + 2</code>，父节点是 <code>(i - 1) /2</code>。</p>
<ol>
<li>首先遍历数组，判断该节点的父节点是否比他小，如果小就交换位置并继续判断，直到他的父节点比他大</li>
<li>重新以上操作 1，直到数组首位是最大值</li>
<li>然后将首位和末尾交换位置并将数组长度减一，表示数组末尾已是最大值，不需要再比较大小</li>
<li>对比左右节点哪个大，然后记住大的节点的索引并且和父节点对比大小，如果子节点大就交换位置</li>
<li>重复以上操作 3 - 4 直到整个数组都是大根堆。</li>
</ol>
<div align="center"><img width="500" class="lazyload inited" data-src="https://user-gold-cdn.xitu.io/2018/4/17/162d2a9ff258dfe1?imageslim" data-width="1280" data-height="367" src="data:image/svg+xml;utf8,<?xml version="1.0"?><svg xmlns="http://www.w3.org/2000/svg" version="1.1" width="1280" height="367"></svg>"></div>
<p>以下是实现该算法的代码</p>
<pre><code class="hljs js" lang="js"><span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">heap</span>(<span class="hljs-params">array</span>) </span>{
<span class="hljs-keyword">if</span> (!checkArray(array)) <span class="hljs-keyword">return</span>
<span class="hljs-comment">// 将最大值交换到首位</span>
<span class="hljs-keyword">for</span> (<span class="hljs-keyword">let</span> i = <span class="hljs-number">0</span>; i < array.length; i++) {
heapInsert(array, i);
}
<span class="hljs-keyword">let</span> size = array.length;
<span class="hljs-comment">// 交换首位和末尾</span>
swap(array, <span class="hljs-number">0</span>, --size);
<span class="hljs-keyword">while</span> (size > <span class="hljs-number">0</span>) {
heapify(array, <span class="hljs-number">0</span>, size);
swap(array, <span class="hljs-number">0</span>, --size);
}
<span class="hljs-keyword">return</span> array;
}
<span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">heapInsert</span>(<span class="hljs-params">array, index</span>) </span>{
<span class="hljs-comment">// 如果当前节点比父节点大，就交换</span>
<span class="hljs-keyword">while</span> (array[index] > array[<span class="hljs-built_in">parseInt</span>((index - <span class="hljs-number">1</span>) / <span class="hljs-number">2</span>)]) {
swap(array, index, <span class="hljs-built_in">parseInt</span>((index - <span class="hljs-number">1</span>) / <span class="hljs-number">2</span>));
<span class="hljs-comment">// 将索引变成父节点</span>
index = <span class="hljs-built_in">parseInt</span>((index - <span class="hljs-number">1</span>) / <span class="hljs-number">2</span>);
}
}
<span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">heapify</span>(<span class="hljs-params">array, index, size</span>) </span>{
<span class="hljs-keyword">let</span> left = index * <span class="hljs-number">2</span> + <span class="hljs-number">1</span>;
<span class="hljs-keyword">while</span> (left < size) {
<span class="hljs-comment">// 判断左右节点大小</span>
<span class="hljs-keyword">let</span> largest =
left + <span class="hljs-number">1</span> < size && array[left] < array[left + <span class="hljs-number">1</span>] ? left + <span class="hljs-number">1</span> : left;
<span class="hljs-comment">// 判断子节点和父节点大小</span>
largest = array[index] < array[largest] ? largest : index;
<span class="hljs-keyword">if</span> (largest === index) <span class="hljs-keyword">break</span>;
swap(array, index, largest);
index = largest;
left = index * <span class="hljs-number">2</span> + <span class="hljs-number">1</span>;
}
}
</code></pre><p>以上代码实现了小根堆，如果需要实现大根堆，只需要把节点对比反一下就好。</p>
<p>该算法的复杂度是 O(logN)</p>
<h3 class="heading" data-id="heading-13">系统自带排序实现</h3>
<p>每个语言的排序内部实现都是不同的。</p>
<p>对于 JS 来说，数组长度大于 10 会采用快排，否则使用插入排序 <a target="_blank" href="https://link.juejin.im/?target=https%3A%2F%2Fgithub.com%2Fv8%2Fv8%2Fblob%2Fad82a40509c5b5b4680d4299c8f08d6c6d31af3c%2Fsrc%2Fjs%2Farray.js%23L760%3A7" rel="nofollow noopener noreferrer">源码实现</a> 。选择插入排序是因为虽然时间复杂度很差，但是在数据量很小的情况下和 <code>O(N * logN)</code>相差无几，然而插入排序需要的常数时间很小，所以相对别的排序来说更快。</p>
<p>对于 Java 来说，还会考虑内部的元素的类型。对于存储对象的数组来说，会采用稳定性好的算法。稳定性的意思就是对于相同值来说，相对顺序不能改变。</p>
<div align="center"><img height="500" class="lazyload inited" data-src="https://user-gold-cdn.xitu.io/2018/4/18/162d7df247dcda00?imageView2/0/w/1280/h/960/format/webp/ignore-error/1" data-width="440" data-height="727" src="data:image/svg+xml;utf8,<?xml version="1.0"?><svg xmlns="http://www.w3.org/2000/svg" version="1.1" width="440" height="727"></svg>"></div>
<h2 class="heading" data-id="heading-14">链表</h2>
<h3 class="heading" data-id="heading-15">反转单向链表</h3>
<p>该题目来自 <a target="_blank" href="https://link.juejin.im/?target=https%3A%2F%2Fleetcode.com%2Fproblems%2Freverse-linked-list%2Fdescription%2F" rel="nofollow noopener noreferrer">LeetCode</a>，题目需要将一个单向链表反转。思路很简单，使用三个变量分别表示当前节点和当前节点的前后节点，虽然这题很简单，但是却是一道面试常考题</p>
<p>以下是实现该算法的代码</p>
<pre><code class="hljs js" lang="js"><span class="hljs-keyword">var</span> reverseList = <span class="hljs-function"><span class="hljs-keyword">function</span>(<span class="hljs-params">head</span>) </span>{
<span class="hljs-comment">// 判断下变量边界问题</span>
<span class="hljs-keyword">if</span> (!head || !head.next) <span class="hljs-keyword">return</span> head
<span class="hljs-comment">// 初始设置为空，因为第一个节点反转后就是尾部，尾部节点指向 null</span>
<span class="hljs-keyword">let</span> pre = <span class="hljs-literal">null</span>
<span class="hljs-keyword">let</span> current = head
<span class="hljs-keyword">let</span> next
<span class="hljs-comment">// 判断当前节点是否为空</span>
<span class="hljs-comment">// 不为空就先获取当前节点的下一节点</span>
<span class="hljs-comment">// 然后把当前节点的 next 设为上一个节点</span>
<span class="hljs-comment">// 然后把 current 设为下一个节点，pre 设为当前节点</span>
<span class="hljs-keyword">while</span>(current) {
next = current.next
current.next = pre
pre = current
current = next
}
<span class="hljs-keyword">return</span> pre
};
</code></pre><h2 class="heading" data-id="heading-16">树</h2>
<h3 class="heading" data-id="heading-17">二叉树的先序，中序，后序遍历</h3>
<p>先序遍历表示先访问根节点，然后访问左节点，最后访问右节点。</p>
<p>中序遍历表示先访问左节点，然后访问根节点，最后访问右节点。</p>
<p>后序遍历表示先访问左节点，然后访问右节点，最后访问根节点。</p>
<h4 class="heading" data-id="heading-18">递归实现</h4>
<p>递归实现相当简单，代码如下</p>
<pre><code class="hljs js" lang="js"><span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">TreeNode</span>(<span class="hljs-params">val</span>) </span>{
<span class="hljs-keyword">this</span>.val = val;
<span class="hljs-keyword">this</span>.left = <span class="hljs-keyword">this</span>.right = <span class="hljs-literal">null</span>;
}
<span class="hljs-keyword">var</span> traversal = <span class="hljs-function"><span class="hljs-keyword">function</span>(<span class="hljs-params">root</span>) </span>{
<span class="hljs-keyword">if</span> (root) {
<span class="hljs-comment">// 先序</span>
<span class="hljs-built_in">console</span>.log(root);
traversal(root.left);
<span class="hljs-comment">// 中序</span>
<span class="hljs-comment">// console.log(root); </span>
traversal(root.right);
<span class="hljs-comment">// 后序</span>
<span class="hljs-comment">// console.log(root);</span>
}
};
</code></pre><p>对于递归的实现来说，只需要理解每个节点都会被访问三次就明白为什么这样实现了。</p>
<h4 class="heading" data-id="heading-19">非递归实现</h4>
<p>非递归实现使用了栈的结构，通过栈的先进后出模拟递归实现。</p>
<p>以下是先序遍历代码实现</p>
<pre><code class="hljs js" lang="js"><span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">pre</span>(<span class="hljs-params">root</span>) </span>{
<span class="hljs-keyword">if</span> (root) {
<span class="hljs-keyword">let</span> stack = [];
<span class="hljs-comment">// 先将根节点 push</span>
stack.push(root);
<span class="hljs-comment">// 判断栈中是否为空</span>
<span class="hljs-keyword">while</span> (stack.length > <span class="hljs-number">0</span>) {
<span class="hljs-comment">// 弹出栈顶元素</span>
root = stack.pop();
<span class="hljs-built_in">console</span>.log(root);
<span class="hljs-comment">// 因为先序遍历是先左后右，栈是先进后出结构</span>
<span class="hljs-comment">// 所以先 push 右边再 push 左边</span>
<span class="hljs-keyword">if</span> (root.right) {
stack.push(root.right);
}
<span class="hljs-keyword">if</span> (root.left) {
stack.push(root.left);
}
}
}
}
</code></pre><p>以下是中序遍历代码实现</p>
<pre><code class="hljs js" lang="js"><span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">mid</span>(<span class="hljs-params">root</span>) </span>{
<span class="hljs-keyword">if</span> (root) {
<span class="hljs-keyword">let</span> stack = [];
<span class="hljs-comment">// 中序遍历是先左再根最后右</span>
<span class="hljs-comment">// 所以首先应该先把最左边节点遍历到底依次 push 进栈</span>
<span class="hljs-comment">// 当左边没有节点时，就打印栈顶元素，然后寻找右节点</span>
<span class="hljs-comment">// 对于最左边的叶节点来说，可以把它看成是两个 null 节点的父节点</span>
<span class="hljs-comment">// 左边打印不出东西就把父节点拿出来打印，然后再看右节点</span>
<span class="hljs-keyword">while</span> (stack.length > <span class="hljs-number">0</span> || root) {
<span class="hljs-keyword">if</span> (root) {
stack.push(root);
root = root.left;
} <span class="hljs-keyword">else</span> {
root = stack.pop();
<span class="hljs-built_in">console</span>.log(root);
root = root.right;
}
}
}
}
</code></pre><p>以下是后序遍历代码实现，该代码使用了两个栈来实现遍历，相比一个栈的遍历来说要容易理解很多</p>
<pre><code class="hljs js" lang="js"><span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">pos</span>(<span class="hljs-params">root</span>) </span>{
<span class="hljs-keyword">if</span> (root) {
<span class="hljs-keyword">let</span> stack1 = [];
<span class="hljs-keyword">let</span> stack2 = [];
<span class="hljs-comment">// 后序遍历是先左再右最后根</span>
<span class="hljs-comment">// 所以对于一个栈来说，应该先 push 根节点</span>
<span class="hljs-comment">// 然后 push 右节点，最后 push 左节点</span>
stack1.push(root);
<span class="hljs-keyword">while</span> (stack1.length > <span class="hljs-number">0</span>) {
root = stack1.pop();
stack2.push(root);
<span class="hljs-keyword">if</span> (root.left) {
stack1.push(root.left);
}
<span class="hljs-keyword">if</span> (root.right) {
stack1.push(root.right);
}
}
<span class="hljs-keyword">while</span> (stack2.length > <span class="hljs-number">0</span>) {
<span class="hljs-built_in">console</span>.log(s2.pop());
}
}
}
</code></pre><h3 class="heading" data-id="heading-20">中序遍历的前驱后继节点</h3>
<p>实现这个算法的前提是节点有一个 <code>parent</code> 的指针指向父节点，根节点指向 <code>null</code> 。</p>
<div align="center"><img width="400" class="lazyload inited" data-src="https://user-gold-cdn.xitu.io/2018/4/24/162f61ad8e8588b7?imageView2/0/w/1280/h/960/format/webp/ignore-error/1" data-width="682" data-height="486" src="data:image/svg+xml;utf8,<?xml version="1.0"?><svg xmlns="http://www.w3.org/2000/svg" version="1.1" width="682" height="486"></svg>"></div>
<p>如图所示，该树的中序遍历结果是 <code>4, 2, 5, 1, 6, 3, 7</code></p>
<h4 class="heading" data-id="heading-21">前驱节点</h4>
<p>对于节点 <code>2</code> 来说，他的前驱节点就是 <code>4</code> ，按照中序遍历原则，可以得出以下结论</p>
<ol>
<li>如果选取的节点的左节点不为空，就找该左节点最右的节点。对于节点 <code>1</code> 来说，他有左节点 <code>2</code> ，那么节点 <code>2</code> 的最右节点就是 <code>5</code></li>
<li>如果左节点为空，且目标节点是父节点的右节点，那么前驱节点为父节点。对于节点 <code>5</code> 来说，没有左节点，且是节点 <code>2</code> 的右节点，所以节点 <code>2</code> 是前驱节点</li>
<li>如果左节点为空，且目标节点是父节点的左节点，向上寻找到第一个是父节点的右节点的节点。对于节点 <code>6</code> 来说，没有左节点，且是节点 <code>3</code> 的左节点，所以向上寻找到节点 <code>1</code> ，发现节点 <code>3</code> 是节点 <code>1</code> 的右节点，所以节点 <code>1</code> 是节点 <code>6</code> 的前驱节点</li>
</ol>
<p>以下是算法实现</p>
<pre><code class="hljs js" lang="js"><span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">predecessor</span>(<span class="hljs-params">node</span>) </span>{
<span class="hljs-keyword">if</span> (!node) <span class="hljs-keyword">return</span>
<span class="hljs-comment">// 结论 1</span>
<span class="hljs-keyword">if</span> (node.left) {
<span class="hljs-keyword">return</span> getRight(node.left)
} <span class="hljs-keyword">else</span> {
<span class="hljs-keyword">let</span> parent = node.parent
<span class="hljs-comment">// 结论 2 3 的判断</span>
<span class="hljs-keyword">while</span>(parent && parent.right === node) {
node = parent
parent = node.parent
}
<span class="hljs-keyword">return</span> parent
}
}
<span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">getRight</span>(<span class="hljs-params">node</span>) </span>{
<span class="hljs-keyword">if</span> (!node) <span class="hljs-keyword">return</span>
node = node.right
<span class="hljs-keyword">while</span>(node) node = node.right
<span class="hljs-keyword">return</span> node
}
</code></pre><h4 class="heading" data-id="heading-22">后继节点</h4>
<p>对于节点 <code>2</code> 来说，他的后继节点就是 <code>5</code> ，按照中序遍历原则，可以得出以下结论</p>
<ol>
<li>如果有右节点，就找到该右节点的最左节点。对于节点 <code>1</code> 来说，他有右节点 <code>3</code> ，那么节点 <code>3</code> 的最左节点就是 <code>6</code></li>
<li>如果没有右节点，就向上遍历直到找到一个节点是父节点的左节点。对于节点 <code>5</code> 来说，没有右节点，就向上寻找到节点 <code>2</code> ，该节点是父节点 <code>1</code> 的左节点，所以节点 <code>1</code> 是后继节点</li>
</ol>
<p>以下是算法实现</p>
<pre><code class="hljs js" lang="js"><span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">successor</span>(<span class="hljs-params">node</span>) </span>{
<span class="hljs-keyword">if</span> (!node) <span class="hljs-keyword">return</span>
<span class="hljs-comment">// 结论 1</span>
<span class="hljs-keyword">if</span> (node.right) {
<span class="hljs-keyword">return</span> getLeft(node.right)
} <span class="hljs-keyword">else</span> {
<span class="hljs-comment">// 结论 2</span>
<span class="hljs-keyword">let</span> parent = node.parent
<span class="hljs-comment">// 判断 parent 为空</span>
<span class="hljs-keyword">while</span>(parent && parent.left === node) {
node = parent
parent = node.parent
}
<span class="hljs-keyword">return</span> parent
}
}
<span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">getLeft</span>(<span class="hljs-params">node</span>) </span>{
<span class="hljs-keyword">if</span> (!node) <span class="hljs-keyword">return</span>
node = node.left
<span class="hljs-keyword">while</span>(node) node = node.left
<span class="hljs-keyword">return</span> node
}
</code></pre><h3 class="heading" data-id="heading-23">树的深度</h3>
<p><strong>树的最大深度</strong>：该题目来自 <a target="_blank" href="https://link.juejin.im/?target=https%3A%2F%2Fleetcode.com%2Fproblems%2Fmaximum-depth-of-binary-tree%2Fdescription%2F" rel="nofollow noopener noreferrer">Leetcode</a>，题目需要求出一颗二叉树的最大深度</p>
<p>以下是算法实现</p>
<pre><code class="hljs js" lang="js"><span class="hljs-keyword">var</span> maxDepth = <span class="hljs-function"><span class="hljs-keyword">function</span>(<span class="hljs-params">root</span>) </span>{
<span class="hljs-keyword">if</span> (!root) <span class="hljs-keyword">return</span> <span class="hljs-number">0</span>
<span class="hljs-keyword">return</span> <span class="hljs-built_in">Math</span>.max(maxDepth(root.left), maxDepth(root.right)) + <span class="hljs-number">1</span>
};
</code></pre><p>对于该递归函数可以这样理解：一旦没有找到节点就会返回 0，每弹出一次递归函数就会加一，树有三层就会得到3。</p>
<h2 class="heading" data-id="heading-24">动态规划</h2>
<p>动态规划背后的基本思想非常简单。就是将一个问题拆分为子问题，一般来说这些子问题都是非常相似的，那么我们可以通过只解决一次每个子问题来达到减少计算量的目的。</p>
<p>一旦得出每个子问题的解，就存储该结果以便下次使用。</p>
<h3 class="heading" data-id="heading-25">斐波那契数列</h3>
<p>斐波那契数列就是从 0 和 1 开始，后面的数都是前两个数之和</p>
<p>0，1，1，2，3，5，8，13，21，34，55，89....</p>
<p>那么显然易见，我们可以通过递归的方式来完成求解斐波那契数列</p>
<pre><code class="hljs js" lang="js"><span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">fib</span>(<span class="hljs-params">n</span>) </span>{
<span class="hljs-keyword">if</span> (n < <span class="hljs-number">2</span> && n >= <span class="hljs-number">0</span>) <span class="hljs-keyword">return</span> n
<span class="hljs-keyword">return</span> fib(n - <span class="hljs-number">1</span>) + fib(n - <span class="hljs-number">2</span>)
}
fib(<span class="hljs-number">10</span>)
</code></pre><p>以上代码已经可以完美的解决问题。但是以上解法却存在很严重的性能问题，当 n 越大的时候，需要的时间是指数增长的，这时候就可以通过动态规划来解决这个问题。</p>
<p>动态规划的本质其实就是两点</p>
<ol>
<li>自底向上分解子问题</li>
<li>通过变量存储已经计算过的解</li>
</ol>
<p>根据上面两点，我们的斐波那契数列的动态规划思路也就出来了</p>
<ol>
<li>斐波那契数列从 0 和 1 开始，那么这就是这个子问题的最底层</li>
<li>通过数组来存储每一位所对应的斐波那契数列的值</li>
</ol>
<pre><code class="hljs js" lang="js"><span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">fib</span>(<span class="hljs-params">n</span>) </span>{
<span class="hljs-keyword">let</span> array = <span class="hljs-keyword">new</span> <span class="hljs-built_in">Array</span>(n + <span class="hljs-number">1</span>).fill(<span class="hljs-literal">null</span>)
array[<span class="hljs-number">0</span>] = <span class="hljs-number">0</span>
array[<span class="hljs-number">1</span>] = <span class="hljs-number">1</span>
<span class="hljs-keyword">for</span> (<span class="hljs-keyword">let</span> i = <span class="hljs-number">2</span>; i <= n; i++) {
array[i] = array[i - <span class="hljs-number">1</span>] + array[i - <span class="hljs-number">2</span>]
}
<span class="hljs-keyword">return</span> array[n]
}
fib(<span class="hljs-number">10</span>)
</code></pre><h3 class="heading" data-id="heading-26">0 - 1背包问题</h3>
<p>该问题可以描述为：给定一组物品，每种物品都有自己的重量和价格，在限定的总重量内，我们如何选择，才能使得物品的总价格最高。每个问题只能放入至多一次。</p>
<p>假设我们有以下物品</p>
<table>
<thead>
<tr>
<th style="text-align:center">物品 ID / 重量</th>
<th style="text-align:center">价值</th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align:center">1</td>
<td style="text-align:center">3</td>
</tr>
<tr>
<td style="text-align:center">2</td>
<td style="text-align:center">7</td>
</tr>
<tr>
<td style="text-align:center">3</td>
<td style="text-align:center">12</td>
</tr>
</tbody>
</table>
<p>对于一个总容量为 5 的背包来说，我们可以放入重量 2 和 3 的物品来达到背包内的物品总价值最高。</p>
<p>对于这个问题来说，子问题就两个，分别是放物品和不放物品，可以通过以下表格来理解子问题</p>
<table>
<thead>
<tr>
<th style="text-align:center">物品 ID / 剩余容量</th>
<th style="text-align:center">0</th>
<th style="text-align:center">1</th>
<th style="text-align:center">2</th>
<th style="text-align:center">3</th>
<th style="text-align:center">4</th>
<th style="text-align:center">5</th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align:center">1</td>
<td style="text-align:center">0</td>
<td style="text-align:center">3</td>
<td style="text-align:center">3</td>
<td style="text-align:center">3</td>
<td style="text-align:center">3</td>
<td style="text-align:center">3</td>
</tr>
<tr>
<td style="text-align:center">2</td>
<td style="text-align:center">0</td>
<td style="text-align:center">3</td>
<td style="text-align:center">7</td>
<td style="text-align:center">10</td>
<td style="text-align:center">10</td>
<td style="text-align:center">10</td>
</tr>
<tr>
<td style="text-align:center">3</td>
<td style="text-align:center">0</td>
<td style="text-align:center">3</td>
<td style="text-align:center">7</td>
<td style="text-align:center">12</td>
<td style="text-align:center">15</td>
<td style="text-align:center">19</td>
</tr>
</tbody>
</table>
<p>直接来分析能放三种物品的情况，也就是最后一行</p>
<ul>
<li>当容量少于 3 时，只取上一行对应的数据，因为当前容量不能容纳物品 3</li>
<li>当容量 为 3 时，考虑两种情况，分别为放入物品 3 和不放物品 3
<ul>
<li>不放物品 3 的情况下，总价值为 10</li>
<li>放入物品 3 的情况下，总价值为 12，所以应该放入物品 3</li>
</ul>
</li>
<li>当容量 为 4 时，考虑两种情况，分别为放入物品 3 和不放物品 3
<ul>
<li>不放物品 3 的情况下，总价值为 10</li>
<li>放入物品 3 的情况下，和放入物品 1 的价值相加，得出总价值为 15，所以应该放入物品 3</li>
</ul>
</li>
<li>当容量 为 5 时，考虑两种情况，分别为放入物品 3 和不放物品 3
<ul>
<li>不放物品 3 的情况下，总价值为 10</li>
<li>放入物品 3 的情况下，和放入物品 2 的价值相加，得出总价值为 19，所以应该放入物品 3</li>
</ul>
</li>
</ul>
<p>以下代码对照上表更容易理解</p>
<pre><code class="hljs js" lang="js"><span class="hljs-comment">/**
* @param {*} w 物品重量
* @param {*} v 物品价值
* @param {*} C 总容量
* @returns
*/</span>
<span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">knapsack</span>(<span class="hljs-params">w, v, C</span>) </span>{
<span class="hljs-keyword">let</span> length = w.length
<span class="hljs-keyword">if</span> (length === <span class="hljs-number">0</span>) <span class="hljs-keyword">return</span> <span class="hljs-number">0</span>
<span class="hljs-comment">// 对照表格，生成的二维数组，第一维代表物品，第二维代表背包剩余容量</span>
<span class="hljs-comment">// 第二维中的元素代表背包物品总价值</span>
<span class="hljs-keyword">let</span> array = <span class="hljs-keyword">new</span> <span class="hljs-built_in">Array</span>(length).fill(<span class="hljs-keyword">new</span> <span class="hljs-built_in">Array</span>(C + <span class="hljs-number">1</span>).fill(<span class="hljs-literal">null</span>))
<span class="hljs-comment">// 完成底部子问题的解</span>
<span class="hljs-keyword">for</span> (<span class="hljs-keyword">let</span> i = <span class="hljs-number">0</span>; i <= C; i++) {
<span class="hljs-comment">// 对照表格第一行， array[0] 代表物品 1</span>
<span class="hljs-comment">// i 代表剩余总容量</span>
<span class="hljs-comment">// 当剩余总容量大于物品 1 的重量时，记录下背包物品总价值，否则价值为 0</span>
array[<span class="hljs-number">0</span>][i] = i >= w[<span class="hljs-number">0</span>] ? v[<span class="hljs-number">0</span>] : <span class="hljs-number">0</span>
}
<span class="hljs-comment">// 自底向上开始解决子问题，从物品 2 开始</span>
<span class="hljs-keyword">for</span> (<span class="hljs-keyword">let</span> i = <span class="hljs-number">1</span>; i < length; i++) {
<span class="hljs-keyword">for</span> (<span class="hljs-keyword">let</span> j = <span class="hljs-number">0</span>; j <= C; j++) {
<span class="hljs-comment">// 这里求解子问题，分别为不放当前物品和放当前物品</span>
<span class="hljs-comment">// 先求不放当前物品的背包总价值，这里的值也就是对应表格中上一行对应的值</span>
array[i][j] = array[i - <span class="hljs-number">1</span>][j]
<span class="hljs-comment">// 判断当前剩余容量是否可以放入当前物品</span>
<span class="hljs-keyword">if</span> (j >= w[i]) {
<span class="hljs-comment">// 可以放入的话，就比大小</span>
<span class="hljs-comment">// 放入当前物品和不放入当前物品，哪个背包总价值大</span>
array[i][j] = <span class="hljs-built_in">Math</span>.max(array[i][j], v[i] + array[i - <span class="hljs-number">1</span>][j - w[i]])
}
}
}
<span class="hljs-keyword">return</span> array[length - <span class="hljs-number">1</span>][C]
}
</code></pre><h3 class="heading" data-id="heading-27">最长递增子序列</h3>
<p>最长递增子序列意思是在一组数字中，找出最长一串递增的数字，比如</p>
<p>0, 3, 4, 17, 2, 8, 6, 10</p>
<p>对于以上这串数字来说，最长递增子序列就是 0, 3, 4, 8, 10，可以通过以下表格更清晰的理解</p>
<table>
<thead>
<tr>
<th style="text-align:center">数字</th>
<th style="text-align:center">0</th>
<th style="text-align:center">3</th>
<th style="text-align:center">4</th>
<th style="text-align:center">17</th>
<th style="text-align:center">2</th>
<th style="text-align:center">8</th>
<th style="text-align:center">6</th>
<th style="text-align:center">10</th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align:center">长度</td>
<td style="text-align:center">1</td>
<td style="text-align:center">2</td>
<td style="text-align:center">3</td>
<td style="text-align:center">4</td>
<td style="text-align:center">2</td>
<td style="text-align:center">4</td>
<td style="text-align:center">4</td>
<td style="text-align:center">5</td>
</tr>
</tbody>
</table>
<p>通过以上表格可以很清晰的发现一个规律，找出刚好比当前数字小的数，并且在小的数组成的长度基础上加一。</p>
<p>这个问题的动态思路解法很简单，直接上代码</p>
<pre><code class="hljs js" lang="js"><span class="hljs-function"><span class="hljs-keyword">function</span> <span class="hljs-title">lis</span>(<span class="hljs-params">n</span>) </span>{
<span class="hljs-keyword">if</span> (n.length === <span class="hljs-number">0</span>) <span class="hljs-keyword">return</span> <span class="hljs-number">0</span>
<span class="hljs-comment">// 创建一个和参数相同大小的数组，并填充值为 1</span>
<span class="hljs-keyword">let</span> array = <span class="hljs-keyword">new</span> <span class="hljs-built_in">Array</span>(n.length).fill(<span class="hljs-number">1</span>)
<span class="hljs-comment">// 从索引 1 开始遍历，因为数组已经所有都填充为 1 了</span>
<span class="hljs-keyword">for</span> (<span class="hljs-keyword">let</span> i = <span class="hljs-number">1</span>; i < n.length; i++) {
<span class="hljs-comment">// 从索引 0 遍历到 i</span>
<span class="hljs-comment">// 判断索引 i 上的值是否大于之前的值</span>
<span class="hljs-keyword">for</span> (<span class="hljs-keyword">let</span> j = <span class="hljs-number">0</span>; j < i; j++) {
<span class="hljs-keyword">if</span> (n[i] > n[j]) {
array[i] = <span class="hljs-built_in">Math</span>.max(array[i], <span class="hljs-number">1</span> + array[j])
}
}
}
<span class="hljs-keyword">let</span> res = <span class="hljs-number">1</span>
<span class="hljs-keyword">for</span> (<span class="hljs-keyword">let</span> i = <span class="hljs-number">0</span>; i < array.length; i++) {
res = <span class="hljs-built_in">Math</span>.max(res, array[i])
}
<span class="hljs-keyword">return</span> res
}
</code></pre></div><section data-v-f8875076="" class="book-comments"><div data-v-f8875076="" class="box-title">留言</div><div data-v-f8875076="" class="comment-box"><div data-v-805e4f50="" data-v-f8875076="" class="comment-form comment-form" id="comment"><div data-v-6795359b="" data-v-67a669a2="" data-v-805e4f50="" data-src="https://mirror-gold-cdn.xitu.io/168e08e4f81e924deaa?imageView2/1/w/100/h/100/q/85/format/webp/interlace/1" class="lazy avatar avatar loaded" style="background-image: url("https://mirror-gold-cdn.xitu.io/168e08e4f81e924deaa?imageView2/1/w/100/h/100/q/85/format/webp/interlace/1");"></div><div data-v-b84335ea="" data-v-805e4f50="" class="auth-card"><div data-v-b84335ea="" class="auth-cover"><span data-v-b84335ea="" class="hint"></span>您需要<a data-v-b84335ea="">绑定手机号</a>后才可在掘金社区内发布内容。</div><textarea data-v-805e4f50="" data-v-b84335ea="" placeholder="评论将在后台进行审核，审核通过后对所有人可见" class="content-input" style="overflow: hidden; overflow-wrap: break-word; height: 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class="collection-arrow-ion ion-arrow-up-b"></i><span data-v-3e0a9536="">0</span></button><span data-v-3e0a9536="" class="sub-comment-btn text-pointer"><span data-v-3e0a9536="" class="title">评论</span><div data-v-3e0a9536="" class="sub-comment--arrow" style="display: none;"></div></span><span data-v-3e0a9536="" class="date">12天前</span><!----></div><!----></div><!----></div></li><li data-v-4c55bf38="" class="item"><div data-v-3e0a9536="" data-v-4c55bf38="" class="comment"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="59f72fa6f265da431c6f9759" class="user-popover-box"><!----><a data-v-3e0a9536="" href="https://juejin.im/user/59f72fa6f265da431c6f9759" target="_blank" rel="" data-v-340b99e1=""><div data-v-6795359b="" data-v-67a669a2="" data-v-3e0a9536="" data-src="https://mirror-gold-cdn.xitu.io/168e08ae3b94b630c5b?imageView2/1/w/100/h/100/q/85/format/webp/interlace/1" class="lazy avatar avatar" style="background-image: none;"></div></a></div><div data-v-3e0a9536="" class="content-box"><div data-v-3e0a9536="" class="header"><div data-v-3e0a9536="" class="user-info"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="59f72fa6f265da431c6f9759" class="user-popover-box"><!----><a data-v-3052f648="" data-v-3e0a9536="" href="https://juejin.im/user/59f72fa6f265da431c6f9759" target="_blank" rel="" class="username" data-v-340b99e1="">sophie旭<!----></a></div><div data-v-3e0a9536="" class="position"></div></div></div><div data-v-3e0a9536="" class="content"><!----><span data-v-3e0a9536="" class="content-html">算法文章：<a href="https://link.juejin.im/?target=https%3A%2F%2Fsegmentfault.com%2Fa%2F1190000010600318%23articleHeader9" target="_blank" rel="nofollow noopener noreferrer">https://segmentfault.com/a/1190000010600318#articleHeader9</a></span><!----></div><div data-v-3e0a9536="" class="footer"><button data-v-3e0a9536="" class="like-btn"><i data-v-3e0a9536="" class="collection-arrow-ion ion-arrow-up-b"></i><span data-v-3e0a9536="">0</span></button><span data-v-3e0a9536="" class="sub-comment-btn text-pointer"><span data-v-3e0a9536="" class="title">评论</span><div data-v-3e0a9536="" class="sub-comment--arrow" style="display: none;"></div></span><span data-v-3e0a9536="" class="date">1月前</span><!----></div><!----></div><!----></div></li><li data-v-4c55bf38="" class="item"><div data-v-3e0a9536="" data-v-4c55bf38="" class="comment"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="5a26acea51882531926e9502" class="user-popover-box"><!----><a data-v-3e0a9536="" href="https://juejin.im/user/5a26acea51882531926e9502" target="_blank" rel="" data-v-340b99e1=""><div data-v-6795359b="" data-v-67a669a2="" data-v-3e0a9536="" data-src="https://user-gold-cdn.xitu.io/2019/1/4/168195919393f6d6?imageView2/1/w/100/h/100/q/85/format/webp/interlace/1" class="lazy avatar avatar" style="background-image: none;"></div></a></div><div data-v-3e0a9536="" class="content-box"><div data-v-3e0a9536="" class="header"><div data-v-3e0a9536="" class="user-info"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="5a26acea51882531926e9502" class="user-popover-box"><!----><a data-v-3052f648="" data-v-3e0a9536="" href="https://juejin.im/user/5a26acea51882531926e9502" target="_blank" rel="" class="username" data-v-340b99e1="">DeCoder<a data-v-7bb9dea1="" data-v-3052f648="" href="https://juejin.im/book/5c90640c5188252d7941f5bb/section/5c9065385188252da6320022" target="_blank" rel="" class="rank"><img data-v-7bb9dea1="" src="./32.常考算法题解析_files/lv-1.636691c.svg" alt="lv-1"></a></a></div><div data-v-3e0a9536="" class="position">求学者 @ 未知</div></div></div><div data-v-3e0a9536="" class="content"><!----><span data-v-3e0a9536="" class="content-html">堆排序的时间复杂度好像写错了。应该是O(n * logn)?</span><!----></div><div data-v-3e0a9536="" class="footer"><button data-v-3e0a9536="" class="like-btn"><i data-v-3e0a9536="" class="collection-arrow-ion ion-arrow-up-b"></i><span data-v-3e0a9536="">0</span></button><span data-v-3e0a9536="" class="sub-comment-btn text-pointer"><span data-v-3e0a9536="" class="title">评论</span><div data-v-3e0a9536="" class="sub-comment--arrow" style="display: none;"></div></span><span data-v-3e0a9536="" class="date">2月前</span><!----></div><!----></div><!----></div></li><li data-v-4c55bf38="" class="item"><div data-v-3e0a9536="" data-v-4c55bf38="" class="comment"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="5a26acea51882531926e9502" class="user-popover-box"><!----><a data-v-3e0a9536="" href="https://juejin.im/user/5a26acea51882531926e9502" target="_blank" rel="" data-v-340b99e1=""><div data-v-6795359b="" data-v-67a669a2="" data-v-3e0a9536="" data-src="https://user-gold-cdn.xitu.io/2019/1/4/168195919393f6d6?imageView2/1/w/100/h/100/q/85/format/webp/interlace/1" class="lazy avatar avatar" style="background-image: none;"></div></a></div><div data-v-3e0a9536="" class="content-box"><div data-v-3e0a9536="" class="header"><div data-v-3e0a9536="" class="user-info"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="5a26acea51882531926e9502" class="user-popover-box"><!----><a data-v-3052f648="" data-v-3e0a9536="" href="https://juejin.im/user/5a26acea51882531926e9502" target="_blank" rel="" class="username" data-v-340b99e1="">DeCoder<a data-v-7bb9dea1="" data-v-3052f648="" href="https://juejin.im/book/5c90640c5188252d7941f5bb/section/5c9065385188252da6320022" target="_blank" rel="" class="rank"><img data-v-7bb9dea1="" src="./32.常考算法题解析_files/lv-1.636691c.svg" alt="lv-1"></a></a></div><div data-v-3e0a9536="" class="position">求学者 @ 未知</div></div></div><div data-v-3e0a9536="" class="content"><!----><span data-v-3e0a9536="" class="content-html collapsed">返回第N小的元素，递归解法：
const partition = (arr, left, right) => {
let i = left, j = right + 1, k = left;
// while(i < j) {
// while(arr[i] < arr[k]) i++;
// while(arr[j] > arr[k]) j--;
// if(i < j) arr.swap(i, j);
// }…</span><span data-v-3e0a9536="" class="show-all">展开全部</span></div><div data-v-3e0a9536="" class="footer"><button data-v-3e0a9536="" class="like-btn"><i data-v-3e0a9536="" class="collection-arrow-ion ion-arrow-up-b"></i><span data-v-3e0a9536="">0</span></button><span data-v-3e0a9536="" class="sub-comment-btn text-pointer"><span data-v-3e0a9536="" class="title">评论</span><div data-v-3e0a9536="" class="sub-comment--arrow" style="display: none;"></div></span><span data-v-3e0a9536="" class="date">2月前</span><!----></div><!----></div><!----></div></li><li data-v-4c55bf38="" class="item"><div data-v-3e0a9536="" data-v-4c55bf38="" class="comment"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="5a26acea51882531926e9502" class="user-popover-box"><!----><a data-v-3e0a9536="" href="https://juejin.im/user/5a26acea51882531926e9502" target="_blank" rel="" data-v-340b99e1=""><div data-v-6795359b="" data-v-67a669a2="" data-v-3e0a9536="" data-src="https://user-gold-cdn.xitu.io/2019/1/4/168195919393f6d6?imageView2/1/w/100/h/100/q/85/format/webp/interlace/1" class="lazy avatar avatar" style="background-image: none;"></div></a></div><div data-v-3e0a9536="" class="content-box"><div data-v-3e0a9536="" class="header"><div data-v-3e0a9536="" class="user-info"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="5a26acea51882531926e9502" class="user-popover-box"><!----><a data-v-3052f648="" data-v-3e0a9536="" href="https://juejin.im/user/5a26acea51882531926e9502" target="_blank" rel="" class="username" data-v-340b99e1="">DeCoder<a data-v-7bb9dea1="" data-v-3052f648="" href="https://juejin.im/book/5c90640c5188252d7941f5bb/section/5c9065385188252da6320022" target="_blank" rel="" class="rank"><img data-v-7bb9dea1="" src="./32.常考算法题解析_files/lv-1.636691c.svg" alt="lv-1"></a></a></div><div data-v-3e0a9536="" class="position">求学者 @ 未知</div></div></div><div data-v-3e0a9536="" class="content"><!----><span data-v-3e0a9536="" class="content-html collapsed">insertionSort貌似有问题？
这是我的：
const insertionSort = arr => {
if(!checkArray(arr)) return;
let i = 0,j = 0;
const currentArr = cloneArray(arr);
for(i = 1; i < currentArr.length; i++) {
tempVal = currentArr[i];…</span><span data-v-3e0a9536="" class="show-all">展开全部</span></div><div data-v-3e0a9536="" class="footer"><button data-v-3e0a9536="" class="like-btn"><i data-v-3e0a9536="" class="collection-arrow-ion ion-arrow-up-b"></i><span data-v-3e0a9536="">0</span></button><span data-v-3e0a9536="" class="sub-comment-btn text-pointer"><span data-v-3e0a9536="" class="title">评论</span><div data-v-3e0a9536="" class="sub-comment--arrow" style="display: none;"></div></span><span data-v-3e0a9536="" class="date">2月前</span><!----></div><!----></div><!----></div></li><li data-v-4c55bf38="" class="item"><div data-v-3e0a9536="" data-v-4c55bf38="" class="comment"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="57ae0cc72e958a0054474d31" class="user-popover-box"><!----><a data-v-3e0a9536="" href="https://juejin.im/user/57ae0cc72e958a0054474d31" target="_blank" rel="" data-v-340b99e1=""><div data-v-6795359b="" data-v-67a669a2="" data-v-3e0a9536="" data-src="https://user-gold-cdn.xitu.io/2016/11/29/9607cb193577eea0bc002af9c3ee9835?imageView2/1/w/100/h/100/q/85/format/webp/interlace/1" class="lazy avatar avatar" style="background-image: none;"></div></a></div><div data-v-3e0a9536="" class="content-box"><div data-v-3e0a9536="" class="header"><div data-v-3e0a9536="" class="user-info"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="57ae0cc72e958a0054474d31" class="user-popover-box"><!----><a data-v-3052f648="" data-v-3e0a9536="" href="https://juejin.im/user/57ae0cc72e958a0054474d31" target="_blank" rel="" class="username" data-v-340b99e1="">o1o<a data-v-7bb9dea1="" data-v-3052f648="" href="https://juejin.im/book/5c90640c5188252d7941f5bb/section/5c9065385188252da6320022" target="_blank" rel="" class="rank"><img data-v-7bb9dea1="" src="./32.常考算法题解析_files/lv-2.f597b88.svg" alt="lv-2"></a></a></div><div data-v-3e0a9536="" class="position">前端</div></div></div><div data-v-3e0a9536="" class="content"><!----><span data-v-3e0a9536="" class="content-html">挣钱也得讲基本法吧，这代码正常跑过吗，看了一个checkarray就不想看了，就算你检查数组的是伪代码，但是在那个函数内部return也太扯淡了吧</span><!----></div><div data-v-3e0a9536="" class="footer"><button data-v-3e0a9536="" class="like-btn"><i data-v-3e0a9536="" class="collection-arrow-ion ion-arrow-up-b"></i><span data-v-3e0a9536="">11</span></button><span data-v-3e0a9536="" class="sub-comment-btn text-pointer"><span data-v-3e0a9536="" class="title">评论</span><div data-v-3e0a9536="" class="sub-comment--arrow" style="display: none;"></div></span><span data-v-3e0a9536="" class="date">3月前</span><!----></div><!----></div><!----></div></li><li data-v-4c55bf38="" class="item"><div data-v-3e0a9536="" data-v-4c55bf38="" class="comment"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="5c3946de51882524ed5b9fc8" class="user-popover-box"><!----><a data-v-3e0a9536="" href="https://juejin.im/user/5c3946de51882524ed5b9fc8" target="_blank" rel="" data-v-340b99e1=""><div data-v-6795359b="" data-v-67a669a2="" data-v-3e0a9536="" data-src="https://mirror-gold-cdn.xitu.io/168e09b93fdc88635b9?imageView2/1/w/100/h/100/q/85/format/webp/interlace/1" class="lazy avatar avatar" style="background-image: none;"></div></a></div><div data-v-3e0a9536="" class="content-box"><div data-v-3e0a9536="" class="header"><div data-v-3e0a9536="" class="user-info"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="5c3946de51882524ed5b9fc8" class="user-popover-box"><!----><a data-v-3052f648="" data-v-3e0a9536="" href="https://juejin.im/user/5c3946de51882524ed5b9fc8" target="_blank" rel="" class="username" data-v-340b99e1="">coma<!----></a></div><div data-v-3e0a9536="" class="position">艺术家</div></div></div><div data-v-3e0a9536="" class="content"><!----><span data-v-3e0a9536="" class="content-html">checkArray 函数返回并不能停止下面代码的执行 应该让 checkArray 返回一个值并在调用他的地方判断他的返回值 再根据结果返回</span><!----></div><div data-v-3e0a9536="" class="footer"><button data-v-3e0a9536="" class="like-btn"><i data-v-3e0a9536="" class="collection-arrow-ion ion-arrow-up-b"></i><span data-v-3e0a9536="">0</span></button><span data-v-3e0a9536="" class="sub-comment-btn text-pointer"><span data-v-3e0a9536="" class="title">收起评论</span><div data-v-3e0a9536="" class="sub-comment--arrow" style=""></div></span><span data-v-3e0a9536="" class="date">3月前</span><!----></div><div data-v-42ee4f35="" data-v-3e0a9536="" class="sub-comment-box sub-comment-box"><div data-v-42ee4f35="" class="sub-comment-box__top"><div data-v-42ee4f35="" class="icon-close"></div><h3 data-v-42ee4f35="">评论</h3></div><div data-v-42ee4f35="" class="sub-comment-box__inner"><div data-v-42ee4f35=""><ul data-v-42ee4f35="" class="sub-comment-list"><li data-v-42ee4f35="" class="item"><div data-v-b9a92622="" data-v-42ee4f35="" class="sub-comment sub-comment"><div data-v-340b99e1="" data-v-b9a92622="" st:block="userPopover" st:state="5c3946de51882524ed5b9fc8" class="user-popover-box user-popover--avatar"><!----><div data-v-6795359b="" data-v-67a669a2="" data-v-b9a92622="" data-src="https://mirror-gold-cdn.xitu.io/168e09b93fdc88635b9?imageView2/1/w/100/h/100/q/85/format/webp/interlace/1" class="lazy avatar avatar" data-v-340b99e1="" style="background-image: none;"></div></div><div data-v-b9a92622="" class="content-box"><div data-v-b9a92622="" class="header"><div data-v-b9a92622="" class="user-info"><div data-v-340b99e1="" data-v-b9a92622="" st:block="userPopover" st:state="5c3946de51882524ed5b9fc8" class="user-popover-box user-popover--info"><!----><a data-v-3052f648="" data-v-b9a92622="" href="https://juejin.im/user/5c3946de51882524ed5b9fc8" target="_blank" rel="" class="username username" data-v-340b99e1="">coma<!----></a></div><div data-v-b9a92622="" class="position">艺术家</div></div></div><div data-v-b9a92622="" class="content"><!----><span data-v-b9a92622="" class="content-html">根据结果决定是否要返回退出排序的函数</span><!----></div><div data-v-b9a92622="" class="footer"><span data-v-b9a92622="" class="date">3月前</span><a data-v-b9a92622="" class="reply-btn">回复</a><!----></div></div><!----></div></li></ul><!----></div><form data-v-42ee4f35="" class="reply-form"><div data-v-b84335ea="" data-v-42ee4f35="" class="auth-card"><div data-v-b84335ea="" class="auth-cover"><span data-v-b84335ea="" class="hint"></span>您需要<a data-v-b84335ea="">绑定手机号</a>后才可在掘金社区内发布内容。</div><textarea data-v-42ee4f35="" data-v-b84335ea="" rows="1" maxlength="1024" placeholder="评论审核通过后显示" style="overflow: hidden; overflow-wrap: break-word; height: 31px;"></textarea></div><button data-v-42ee4f35="" class="reply-btn">评论</button></form></div></div></div><!----></div></li><li data-v-4c55bf38="" class="item"><div data-v-3e0a9536="" data-v-4c55bf38="" class="comment"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="5c421123f265da614933e8d7" class="user-popover-box"><!----><a data-v-3e0a9536="" href="https://juejin.im/user/5c421123f265da614933e8d7" target="_blank" rel="" data-v-340b99e1=""><div data-v-6795359b="" data-v-67a669a2="" data-v-3e0a9536="" data-src="https://mirror-gold-cdn.xitu.io/168e09bc93b83a9f107?imageView2/1/w/100/h/100/q/85/format/webp/interlace/1" class="lazy avatar avatar" style="background-image: none;"></div></a></div><div data-v-3e0a9536="" class="content-box"><div data-v-3e0a9536="" class="header"><div data-v-3e0a9536="" class="user-info"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="5c421123f265da614933e8d7" class="user-popover-box"><!----><a data-v-3052f648="" data-v-3e0a9536="" href="https://juejin.im/user/5c421123f265da614933e8d7" target="_blank" rel="" class="username" data-v-340b99e1="">诺维阿鑫😊<!----></a></div><div data-v-3e0a9536="" class="position"></div></div></div><div data-v-3e0a9536="" class="content"><!----><span data-v-3e0a9536="" class="content-html">那道用快排的思维求第K个大值的 请问less的作用是什么</span><!----></div><div data-v-3e0a9536="" class="footer"><button data-v-3e0a9536="" class="like-btn"><i data-v-3e0a9536="" class="collection-arrow-ion ion-arrow-up-b"></i><span data-v-3e0a9536="">0</span></button><span data-v-3e0a9536="" class="sub-comment-btn text-pointer"><span data-v-3e0a9536="" class="title">评论</span><div data-v-3e0a9536="" class="sub-comment--arrow" style="display: none;"></div></span><span data-v-3e0a9536="" class="date">3月前</span><!----></div><!----></div><!----></div></li><li data-v-4c55bf38="" class="item"><div data-v-3e0a9536="" data-v-4c55bf38="" class="comment"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="5a9e3e21518825558a061ceb" class="user-popover-box"><!----><a data-v-3e0a9536="" href="https://juejin.im/user/5a9e3e21518825558a061ceb" target="_blank" rel="" data-v-340b99e1=""><div data-v-6795359b="" data-v-67a669a2="" data-v-3e0a9536="" data-src="https://mirror-gold-cdn.xitu.io/168e08e44b4d909b1e6?imageView2/1/w/100/h/100/q/85/format/webp/interlace/1" class="lazy avatar avatar" style="background-image: none;"></div></a></div><div data-v-3e0a9536="" class="content-box"><div data-v-3e0a9536="" class="header"><div data-v-3e0a9536="" class="user-info"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="5a9e3e21518825558a061ceb" class="user-popover-box"><!----><a data-v-3052f648="" data-v-3e0a9536="" href="https://juejin.im/user/5a9e3e21518825558a061ceb" target="_blank" rel="" class="username" data-v-340b99e1="">没想到你是这样的青年<!----></a></div><div data-v-3e0a9536="" class="position"></div></div></div><div data-v-3e0a9536="" class="content"><!----><span data-v-3e0a9536="" class="content-html">学算法最大的问题就是，学会了，会写了，长时间不用就忘了</span><!----></div><div data-v-3e0a9536="" class="footer"><button data-v-3e0a9536="" class="like-btn"><i data-v-3e0a9536="" class="collection-arrow-ion ion-arrow-up-b"></i><span data-v-3e0a9536="">4</span></button><span data-v-3e0a9536="" class="sub-comment-btn text-pointer"><span data-v-3e0a9536="" class="title">评论</span><div data-v-3e0a9536="" class="sub-comment--arrow" style="display: none;"></div></span><span data-v-3e0a9536="" class="date">3月前</span><!----></div><!----></div><!----></div></li><li data-v-4c55bf38="" class="item"><div data-v-3e0a9536="" data-v-4c55bf38="" class="comment"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="585385e91b69e6006c8569fc" class="user-popover-box"><!----><a data-v-3e0a9536="" href="https://juejin.im/user/585385e91b69e6006c8569fc" target="_blank" rel="" data-v-340b99e1=""><div data-v-6795359b="" data-v-67a669a2="" data-v-3e0a9536="" data-src="https://b-gold-cdn.xitu.io/v3/static/img/default-avatar.e30559a.svg" class="lazy avatar avatar" style="background-image: none;"></div></a></div><div data-v-3e0a9536="" class="content-box"><div data-v-3e0a9536="" class="header"><div data-v-3e0a9536="" class="user-info"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="585385e91b69e6006c8569fc" class="user-popover-box"><!----><a data-v-3052f648="" data-v-3e0a9536="" href="https://juejin.im/user/585385e91b69e6006c8569fc" target="_blank" rel="" class="username" data-v-340b99e1="">火华社的怪兽<!----></a></div><div data-v-3e0a9536="" class="position"></div></div></div><div data-v-3e0a9536="" class="content"><!----><span data-v-3e0a9536="" class="content-html">使用quickSort()当有多个相同值，这个快排有bug</span><!----></div><div data-v-3e0a9536="" class="footer"><button data-v-3e0a9536="" class="like-btn"><i data-v-3e0a9536="" class="collection-arrow-ion ion-arrow-up-b"></i><span data-v-3e0a9536="">2</span></button><span data-v-3e0a9536="" class="sub-comment-btn text-pointer"><span data-v-3e0a9536="" class="title">评论</span><div data-v-3e0a9536="" class="sub-comment--arrow" style="display: none;"></div></span><span data-v-3e0a9536="" class="date">3月前</span><!----></div><!----></div><!----></div></li><li data-v-4c55bf38="" class="item"><div data-v-3e0a9536="" data-v-4c55bf38="" class="comment"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="5ba9f38ce51d450e8477bd7a" class="user-popover-box"><!----><a data-v-3e0a9536="" href="https://juejin.im/user/5ba9f38ce51d450e8477bd7a" target="_blank" rel="" data-v-340b99e1=""><div data-v-6795359b="" data-v-67a669a2="" data-v-3e0a9536="" data-src="https://mirror-gold-cdn.xitu.io/168e094945d95739467?imageView2/1/w/100/h/100/q/85/format/webp/interlace/1" class="lazy avatar avatar" style="background-image: none;"></div></a></div><div data-v-3e0a9536="" class="content-box"><div data-v-3e0a9536="" class="header"><div data-v-3e0a9536="" class="user-info"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="5ba9f38ce51d450e8477bd7a" class="user-popover-box"><!----><a data-v-3052f648="" data-v-3e0a9536="" href="https://juejin.im/user/5ba9f38ce51d450e8477bd7a" target="_blank" rel="" class="username" data-v-340b99e1="">yeyan1996<a data-v-7bb9dea1="" data-v-3052f648="" href="https://juejin.im/book/5c90640c5188252d7941f5bb/section/5c9065385188252da6320022" target="_blank" rel="" class="rank"><img data-v-7bb9dea1="" src="./32.常考算法题解析_files/lv-4.2c3fafd.svg" alt="lv-4"></a></a></div><div data-v-3e0a9536="" class="position">FE @ 找个上海的互联网企业~</div></div></div><div data-v-3e0a9536="" class="content"><!----><span data-v-3e0a9536="" class="content-html collapsed">function quickSort(array, left, right) {
if (left < right) {
swap(array, , right)
// 随机取值，然后和末尾交换，这样做比固定取一个位置的复杂度略低
let indexs = part(array, parseInt(Math.random() * (right - left + 1)) + left, right);
quickSort(array, left, indexs[0]);
quickSort(array, indexs[1] + 1, right);…</span><span data-v-3e0a9536="" class="show-all">展开全部</span></div><div data-v-3e0a9536="" class="footer"><button data-v-3e0a9536="" class="like-btn"><i data-v-3e0a9536="" class="collection-arrow-ion ion-arrow-up-b"></i><span data-v-3e0a9536="">3</span></button><span data-v-3e0a9536="" class="sub-comment-btn text-pointer"><span data-v-3e0a9536="" class="title">评论</span><div data-v-3e0a9536="" class="sub-comment--arrow" style="display: none;"></div></span><span data-v-3e0a9536="" class="date">3月前</span><!----></div><!----></div><!----></div></li><li data-v-4c55bf38="" class="item"><div data-v-3e0a9536="" data-v-4c55bf38="" class="comment"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="5971c4bff265da6c375ac767" class="user-popover-box"><!----><a data-v-3e0a9536="" href="https://juejin.im/user/5971c4bff265da6c375ac767" target="_blank" rel="" data-v-340b99e1=""><div data-v-6795359b="" data-v-67a669a2="" data-v-3e0a9536="" data-src="https://mirror-gold-cdn.xitu.io/168e0869ed791d6fc60?imageView2/1/w/100/h/100/q/85/format/webp/interlace/1" class="lazy avatar avatar" style="background-image: none;"></div></a></div><div data-v-3e0a9536="" class="content-box"><div data-v-3e0a9536="" class="header"><div data-v-3e0a9536="" class="user-info"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="5971c4bff265da6c375ac767" class="user-popover-box"><!----><a data-v-3052f648="" data-v-3e0a9536="" href="https://juejin.im/user/5971c4bff265da6c375ac767" target="_blank" rel="" class="username" data-v-340b99e1="">shability<!----></a></div><div data-v-3e0a9536="" class="position"></div></div></div><div data-v-3e0a9536="" class="content"><!----><span data-v-3e0a9536="" class="content-html">function checkArray(array) {
if (!array) return
}
这个根本判断不出来是否是数组</span><!----></div><div data-v-3e0a9536="" class="footer"><button data-v-3e0a9536="" class="like-btn"><i data-v-3e0a9536="" class="collection-arrow-ion ion-arrow-up-b"></i><span data-v-3e0a9536="">1</span></button><span data-v-3e0a9536="" class="sub-comment-btn text-pointer"><span data-v-3e0a9536="" class="title">评论</span><div data-v-3e0a9536="" class="sub-comment--arrow" style="display: none;"></div></span><span data-v-3e0a9536="" class="date">3月前</span><!----></div><!----></div><!----></div></li><li data-v-4c55bf38="" class="item"><div data-v-3e0a9536="" data-v-4c55bf38="" class="comment"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="58498d34b123db006607eb30" class="user-popover-box"><!----><a data-v-3e0a9536="" href="https://juejin.im/user/58498d34b123db006607eb30" target="_blank" rel="" data-v-340b99e1=""><div data-v-6795359b="" data-v-67a669a2="" data-v-3e0a9536="" data-src="https://user-gold-cdn.xitu.io/2019/3/31/169d3ac880701bd6?imageView2/1/w/100/h/100/q/85/format/webp/interlace/1" class="lazy avatar avatar" style="background-image: none;"></div></a></div><div data-v-3e0a9536="" class="content-box"><div data-v-3e0a9536="" class="header"><div data-v-3e0a9536="" class="user-info"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="58498d34b123db006607eb30" class="user-popover-box"><!----><a data-v-3052f648="" data-v-3e0a9536="" href="https://juejin.im/user/58498d34b123db006607eb30" target="_blank" rel="" class="username" data-v-340b99e1="">星辰大海1481215283747<!----></a></div><div data-v-3e0a9536="" class="position">前端</div></div></div><div data-v-3e0a9536="" class="content"><!----><span data-v-3e0a9536="" class="content-html">有点多，好多用不上</span><!----></div><div data-v-3e0a9536="" class="footer"><button data-v-3e0a9536="" class="like-btn"><i data-v-3e0a9536="" class="collection-arrow-ion ion-arrow-up-b"></i><span data-v-3e0a9536="">0</span></button><span data-v-3e0a9536="" class="sub-comment-btn text-pointer"><span data-v-3e0a9536="" class="title">收起评论</span><div data-v-3e0a9536="" class="sub-comment--arrow" style=""></div></span><span data-v-3e0a9536="" class="date">3月前</span><!----></div><div data-v-42ee4f35="" data-v-3e0a9536="" class="sub-comment-box sub-comment-box"><div data-v-42ee4f35="" class="sub-comment-box__top"><div data-v-42ee4f35="" class="icon-close"></div><h3 data-v-42ee4f35="">评论</h3></div><div data-v-42ee4f35="" class="sub-comment-box__inner"><div data-v-42ee4f35=""><ul data-v-42ee4f35="" class="sub-comment-list"><li data-v-42ee4f35="" class="item"><div data-v-b9a92622="" data-v-42ee4f35="" class="sub-comment sub-comment"><div data-v-340b99e1="" data-v-b9a92622="" st:block="userPopover" st:state="58498d34b123db006607eb30" class="user-popover-box user-popover--avatar"><!----><div data-v-6795359b="" data-v-67a669a2="" data-v-b9a92622="" data-src="https://user-gold-cdn.xitu.io/2019/3/31/169d3ac880701bd6?imageView2/1/w/100/h/100/q/85/format/webp/interlace/1" class="lazy avatar avatar" data-v-340b99e1="" style="background-image: none;"></div></div><div data-v-b9a92622="" class="content-box"><div data-v-b9a92622="" class="header"><div data-v-b9a92622="" class="user-info"><div data-v-340b99e1="" data-v-b9a92622="" st:block="userPopover" st:state="58498d34b123db006607eb30" class="user-popover-box user-popover--info"><!----><a data-v-3052f648="" data-v-b9a92622="" href="https://juejin.im/user/58498d34b123db006607eb30" target="_blank" rel="" class="username username" data-v-340b99e1="">星辰大海1481215283747<!----></a></div><div data-v-b9a92622="" class="position">前端</div></div></div><div data-v-b9a92622="" class="content"><!----><span data-v-b9a92622="" class="content-html">这是我</span><!----></div><div data-v-b9a92622="" class="footer"><span data-v-b9a92622="" class="date">2月前</span><a data-v-b9a92622="" class="reply-btn">回复</a><!----></div></div><!----></div></li></ul><!----></div><form data-v-42ee4f35="" class="reply-form"><div data-v-b84335ea="" data-v-42ee4f35="" class="auth-card"><div data-v-b84335ea="" class="auth-cover"><span data-v-b84335ea="" class="hint"></span>您需要<a data-v-b84335ea="">绑定手机号</a>后才可在掘金社区内发布内容。</div><textarea data-v-42ee4f35="" data-v-b84335ea="" rows="1" maxlength="1024" placeholder="评论审核通过后显示" style="overflow: hidden; overflow-wrap: break-word; height: 31px;"></textarea></div><button data-v-42ee4f35="" class="reply-btn">评论</button></form></div></div></div><!----></div></li><li data-v-4c55bf38="" class="item"><div data-v-3e0a9536="" data-v-4c55bf38="" class="comment"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="5a20471551882529642179dd" class="user-popover-box"><!----><a data-v-3e0a9536="" href="https://juejin.im/user/5a20471551882529642179dd" target="_blank" rel="" data-v-340b99e1=""><div data-v-6795359b="" data-v-67a669a2="" data-v-3e0a9536="" data-src="https://mirror-gold-cdn.xitu.io/168e08c1aa96b835d2b?imageView2/1/w/100/h/100/q/85/format/webp/interlace/1" class="lazy avatar avatar" style="background-image: none;"></div></a></div><div data-v-3e0a9536="" class="content-box"><div data-v-3e0a9536="" class="header"><div data-v-3e0a9536="" class="user-info"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="5a20471551882529642179dd" class="user-popover-box"><!----><a data-v-3052f648="" data-v-3e0a9536="" href="https://juejin.im/user/5a20471551882529642179dd" target="_blank" rel="" class="username" data-v-340b99e1="">ForinArray<!----></a></div><div data-v-3e0a9536="" class="position"></div></div></div><div data-v-3e0a9536="" class="content"><!----><span data-v-3e0a9536="" class="content-html">重新评论下那个getRight 如果我没理解错的话 这个永远都会返回null这种的吧，等到return的时候 node这个变量已经变成null了啊 是不是应该修改成这样 function getRight(node) {
while (node) {
if (!node.right) {
return node
}
node = node.right
}
}</span><!----></div><div data-v-3e0a9536="" class="footer"><button data-v-3e0a9536="" class="like-btn"><i data-v-3e0a9536="" class="collection-arrow-ion ion-arrow-up-b"></i><span data-v-3e0a9536="">0</span></button><span data-v-3e0a9536="" class="sub-comment-btn text-pointer"><span data-v-3e0a9536="" class="title">评论</span><div data-v-3e0a9536="" class="sub-comment--arrow" style="display: none;"></div></span><span data-v-3e0a9536="" class="date">3月前</span><!----></div><!----></div><!----></div></li><li data-v-4c55bf38="" class="item"><div data-v-3e0a9536="" data-v-4c55bf38="" class="comment"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="5a20471551882529642179dd" class="user-popover-box"><!----><a data-v-3e0a9536="" href="https://juejin.im/user/5a20471551882529642179dd" target="_blank" rel="" data-v-340b99e1=""><div data-v-6795359b="" data-v-67a669a2="" data-v-3e0a9536="" data-src="https://mirror-gold-cdn.xitu.io/168e08c1aa96b835d2b?imageView2/1/w/100/h/100/q/85/format/webp/interlace/1" class="lazy avatar avatar" style="background-image: none;"></div></a></div><div data-v-3e0a9536="" class="content-box"><div data-v-3e0a9536="" class="header"><div data-v-3e0a9536="" class="user-info"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="5a20471551882529642179dd" class="user-popover-box"><!----><a data-v-3052f648="" data-v-3e0a9536="" href="https://juejin.im/user/5a20471551882529642179dd" target="_blank" rel="" class="username" data-v-340b99e1="">ForinArray<!----></a></div><div data-v-3e0a9536="" class="position"></div></div></div><div data-v-3e0a9536="" class="content"><!----><span data-v-3e0a9536="" class="content-html">前驱节点getRight哪里 第一行是不是应该改成这样 if(!node.right) return node
因为调用前有if(node.left)的判断 所以node必然存在，但是这个节点不一定存在right节点</span><!----></div><div data-v-3e0a9536="" class="footer"><button data-v-3e0a9536="" class="like-btn"><i data-v-3e0a9536="" class="collection-arrow-ion ion-arrow-up-b"></i><span data-v-3e0a9536="">0</span></button><span data-v-3e0a9536="" class="sub-comment-btn text-pointer"><span data-v-3e0a9536="" class="title">评论</span><div data-v-3e0a9536="" class="sub-comment--arrow" style="display: none;"></div></span><span data-v-3e0a9536="" class="date">3月前</span><!----></div><!----></div><!----></div></li><li data-v-4c55bf38="" class="item"><div data-v-3e0a9536="" data-v-4c55bf38="" class="comment"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="5796cf5e128fe10056cb8eaa" class="user-popover-box"><!----><a data-v-3e0a9536="" href="https://juejin.im/user/5796cf5e128fe10056cb8eaa" target="_blank" rel="" data-v-340b99e1=""><div data-v-6795359b="" data-v-67a669a2="" data-v-3e0a9536="" data-src="https://user-gold-cdn.xitu.io/2016/11/29/8d35d756d09033e27b1b1264fa4578bc?imageView2/1/w/100/h/100/q/85/format/webp/interlace/1" class="lazy avatar avatar" style="background-image: none;"></div></a></div><div data-v-3e0a9536="" class="content-box"><div data-v-3e0a9536="" class="header"><div data-v-3e0a9536="" class="user-info"><div data-v-340b99e1="" data-v-3e0a9536="" st:block="userPopover" st:state="5796cf5e128fe10056cb8eaa" class="user-popover-box"><!----><a data-v-3052f648="" data-v-3e0a9536="" href="https://juejin.im/user/5796cf5e128fe10056cb8eaa" target="_blank" rel="" class="username" data-v-340b99e1="">石子路<!----></a></div><div data-v-3e0a9536="" class="position">视觉体验还原师 @ 滴滴出行</div></div></div><div data-v-3e0a9536="" class="content"><!----><span data-v-3e0a9536="" class="content-html">厉害了！</span><!----></div><div data-v-3e0a9536="" class="footer"><button data-v-3e0a9536="" class="like-btn"><i data-v-3e0a9536="" class="collection-arrow-ion ion-arrow-up-b"></i><span data-v-3e0a9536="">0</span></button><span data-v-3e0a9536="" class="sub-comment-btn text-pointer"><span data-v-3e0a9536="" class="title">评论</span><div data-v-3e0a9536="" class="sub-comment--arrow" style="display: none;"></div></span><span data-v-3e0a9536="" class="date">3月前</span><!----></div><!----></div><!----></div></li><!----></ul></section></div></div><!----><!----></div></div><div data-v-5cee32d6="" data-v-54b8e7d6="" class="book-handle book-direction"><div data-v-5cee32d6="" class="step-btn step-btn--prev"><img data-v-5cee32d6="" src="./32.常考算法题解析_files/prev.87ad47e.svg"></div><div data-v-5cee32d6="" class="step-btn step-btn--next"><img data-v-5cee32d6="" src="./32.常考算法题解析_files/next.54d8a35.svg"></div><!----><!----></div><!----></div></div></section><!----><!----><!----><div data-v-cd2ca6a6="" data-v-54b8e7d6="" class="image-viewer-box"><!----></div></div><!----></div>
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