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<h2 class="wrapper title">0000.为什么需要复杂度分析</h2>
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<span>Author：</span><span>flypython</span> | <span>Date: </span><span>2020-02-24</span> | <span>Category：</span><span><a href="/fly/飞蟒微课堂/" title="飞蟒微课堂">飞蟒微课堂</a><a href="/fly/飞蟒微课堂/LeetCode/" title="LeetCode">LeetCode</a></span>
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<h4 id="写在前面"><a href="#写在前面" class="headerlink" title="写在前面"></a>写在前面</h4><p>FlyPython推出的《Python面试专项课程》原计划在春节假期后开始更新，考虑到后面刷题的需求可能增多，我们对题目进行了调整并于今天开始更新。</p>
<p>原先的<a href="https://mp.weixin.qq.com/s/8EtaSfjyWWJikumG3_l_YQ" target="_blank" rel="noopener">LeetCode刷题计划</a>中的精选TOP面试题推后，我们先过一遍《程序员面试金典》这本书上的面试题，为基础薄弱的同学夯实基础。</p>
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<li>网址：<a href="https://leetcode-cn.com/problemset/lcci/" target="_blank" rel="noopener">https://leetcode-cn.com/problemset/lcci/</a></li>
<li>总题数： 97</li>
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<p>欢迎关注我们的公众号flypython和网站flypython.com，我们将持续更新这个专项课程。</p>
<h2 id="为什么需要复杂度分析"><a href="#为什么需要复杂度分析" class="headerlink" title="为什么需要复杂度分析"></a>为什么需要复杂度分析</h2><p>今天是这个专项课程的第0课，我们带来算法学习最重要的一个部分：复杂度分析。</p>
<p>为什么说复杂度分析是最重要呢？因为数据结构和算法其实解决的是”快”和”省”的问题。”快”就是如何让代码运行得更快，”省”就是让代码节约存储空间。怎么样来衡量你编写的算法代码执行效率呢？这里引出来今天要讲的内容：时间、空间复杂度分析。</p>
<p>你可能会说，不就是执行效率么？我把代码跑一遍，增加cProfile的一些操作，不就得到了执行时间和内存占用么。</p>
<p>其实你上面说的也是一种方法，叫事后统计法，在做性能优化时会大量采用。但是事后统计法有它的局限性。第一，它依赖测试环境，测试环境不一样，得到的数据会都不一样；第二，它受数据规模的影响大，比如排序，排10个数和排100万个数肯定不一样。所以我们不能以具体测试数据来表示，需要的是一个估算的执行效率的方法。这个方法就是我们要讲的大O复杂度表示法。</p>
<h2 id="大O复杂度表示法"><a href="#大O复杂度表示法" class="headerlink" title="大O复杂度表示法"></a>大O复杂度表示法</h2><p>我们从一个例子开始，下面是求1,2,3,…n的累加和的代码</p>
<figure class="highlight plain"><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></pre></td><td class="code"><pre><span class="line">1 int cal(int n) &#123;</span><br><span class="line">2 int sum = 0;</span><br><span class="line">3 int i = 1;</span><br><span class="line">4 for (; i &lt;= n; ++i) &#123;</span><br><span class="line">5 sum = sum + i;</span><br><span class="line">6 &#125;</span><br><span class="line">7 return sum;</span><br><span class="line">8 &#125;</span><br></pre></td></tr></table></figure>
<p>我们假设每一行代码的执行时间是相同的，为unit_time，那么这一段代码，第2，3行执行了一次，4，5行执行了n遍，第7行执行了一次，那总共执行的次数2n+3,执行时间为(2n+3)* unit_time。</p>
<p>可以看出，所有代码的执行时间 T(n) 与每行代码的执行次数成正比。可以用以下公式表示：</p>
<p><img src="https://tva1.sinaimg.cn/large/0082zybply1gc7k3lrcljj30vh031dfr.jpg" alt></p>
<p>其中T(n)为所有代码的执行时间<br>其中f(n)表示每行代码执行的次数总和</p>
<p>公式中的O，表示代码的执行时间 T(n) 与 f(n) 表达式成正比。我们的例子中，T(n)=O(2n+3)，这就是大O时间复杂度表示法。</p>
<p>大 O 时间复杂度实际上并不具体表示代码真正的执行时间，而是表示代码执行时间随数据规模增长的变化趋势，所以，也叫作渐进时间复杂度（asymptotic time complexity），简称时间复杂度。</p>
<h4 id="时间复杂度分析"><a href="#时间复杂度分析" class="headerlink" title="时间复杂度分析"></a>时间复杂度分析</h4><ul>
<li>O(1): 常数</li>
<li>O(logn): 对数</li>
<li>O(n): 线性</li>
<li>O(nlogn)：线性对数</li>
<li>O(n^2): 平⽅</li>
<li>O(n^3): 立⽅ </li>
<li>O(n^k): K次方</li>
<li>O(2^n): 指数 </li>
<li>O(n!): 阶乘 </li>
</ul>
<h5 id="O-1"><a href="#O-1" class="headerlink" title="O(1)"></a>O(1)</h5><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">int a=0;</span><br><span class="line">int b=a;</span><br></pre></td></tr></table></figure>
<p>O(1) 只是常量级时间复杂度的一种表示方法，并不是指只执行了一行代码。只要代码的执行时间不随 n 的增大而增长，这样代码的时间复杂度我们都记作 O(1)。</p>
<h5 id="O-log-n"><a href="#O-log-n" class="headerlink" title="O(log n)"></a>O(log n)</h5><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">for(int i=1; i&lt;n; i=i*2)&#123;</span><br><span class="line"> printf(i);</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<p>对数阶时间复杂度非常常见，同时也是最难分析的一种时间复杂度。从代码中可以看出，变量 i 的值从 1 开始取，每循环一次就乘以 2。当大于 n 时，循环结束。还记得我们高中学过的等比数列吗？实际上，变量 i 的取值就是一个等比数列</p>
<p><img src="https://tva1.sinaimg.cn/large/0082zybply1gc7k3xx5grj30vq043aa2.jpg" alt></p>
<p>通过 2^x = n 求解 x, x=log2n，这段代码的时间复杂度就是 O(log2n)，忽略系数，表示为O(logn)。</p>
<h5 id="O-n"><a href="#O-n" class="headerlink" title="O(n)"></a>O(n)</h5><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">for(int i=1; i&lt;=n; i++)&#123;</span><br><span class="line"> printf(i);</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<p>for循环代码执行了n遍，因此它消耗的时间是随着n的变化而变化的，所以通过O(n)来表示。如果平行存在一个for循环m遍的，如下:</p>
<figure class="highlight plain"><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></pre></td><td class="code"><pre><span class="line">for(int i=1; i&lt;=n; i++)&#123;</span><br><span class="line"> printf(i);</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line">for(int i=1; i&lt;=m; i++)&#123;</span><br><span class="line"> printf(i);</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<p>那时间复杂度为O(m+n)。</p>
<h5 id="O-nlogn"><a href="#O-nlogn" class="headerlink" title="O(nlogn)"></a>O(nlogn)</h5><figure class="highlight plain"><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></pre></td><td class="code"><pre><span class="line">for(int i=1; i&lt;=n; i++)&#123;</span><br><span class="line"> for(int j=1; j&lt;=n;j=j*2) &#123;</span><br><span class="line"> printf(i,j);</span><br><span class="line"> &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<p>线性对数O(nlogn)就是将时间复杂度为O(logn)的代码循环N遍的话，那么它的时间复杂度就是n* O(logn)，也就是了O(nlogn)</p>
<h5 id="O-n-2"><a href="#O-n-2" class="headerlink" title="O(n^2)"></a>O(n^2)</h5><figure class="highlight plain"><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></pre></td><td class="code"><pre><span class="line">for(int i=1; i&lt;=n; i++)&#123;</span><br><span class="line"> for(int j=1; j&lt;=n;j++) &#123;</span><br><span class="line"> printf(i,j);</span><br><span class="line"> &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<p>平方就是O(n)代码再嵌套循环一遍，2层n循环，O(n * n) 为O(n^2)。如果嵌套3层n循环，则为O(n^3)，如果嵌套k层，则为O(n^k)。</p>
<p>如果其中一层循环中的n改为m，如下所示:</p>
<figure class="highlight plain"><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></pre></td><td class="code"><pre><span class="line">for(int i=1; i&lt;=m; i++)&#123;</span><br><span class="line"> for(int j=1; j&lt;=n;j++) &#123;</span><br><span class="line"> printf(i,j);</span><br><span class="line"> &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<p>那它的时间复杂度为O(m * n)</p>
<h5 id="O-2-n-和O-n"><a href="#O-2-n-和O-n" class="headerlink" title="O(2^n)和O(n!)"></a>O(2^n)和O(n!)</h5><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">for(int i=1; i&lt;=math.pow(2,n); i++)&#123;</span><br><span class="line"> printf(i);</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">for(int i=1; i&lt;=factorial(n); i++)&#123;</span><br><span class="line"> printf(i);</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<p>以上两个代码例子分别为O(2^n)复杂度和O(n!)复杂度，指数复杂度和阶乘复杂度属于非多项式量级，我们把时间复杂度为非多项式量级的算法问题叫作 NP（Non-Deterministic Polynomial，非确定多项式）问题。</p>
<p>当数据规模 n 越来越大时，非多项式量级算法的执行时间会急剧增加，求解问题的执行时间会无限增长。所以，非多项式时间复杂度的算法其实是非常低效的算法。</p>
<p>具体增长曲线如下图所示：</p>
<p><img src="https://tva1.sinaimg.cn/large/0082zybply1gc7k46ndw2j319m0u00yx.jpg" alt></p>
<h4 id="空间复杂度分析"><a href="#空间复杂度分析" class="headerlink" title="空间复杂度分析"></a>空间复杂度分析</h4><ul>
<li>O(1)</li>
<li>O(n)</li>
<li>O(n²)</li>
</ul>
<h5 id="O-1-1"><a href="#O-1-1" class="headerlink" title="O(1)"></a>O(1)</h5><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">int i=0;</span><br><span class="line">int j=1;</span><br><span class="line">int m = i + j;</span><br></pre></td></tr></table></figure>
<p>代码中的i,j,m不随着处理数据量变化而变化，算法需要的临时空间是固定的，可以表示为空间复杂度O(1)</p>
<h5 id="O-n-1"><a href="#O-n-1" class="headerlink" title="O(n)"></a>O(n)</h5><figure class="highlight plain"><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></pre></td><td class="code"><pre><span class="line">int[] m = new int[n];</span><br><span class="line">for(i=1; i&lt;=n; ++i)</span><br><span class="line">&#123;</span><br><span class="line"> printf(i);</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<p>这段代码中，第一行new了一个数组出来，这个数据占用的大小为n，这段代码的2-5行，虽然有循环，但没有再分配新的空间，所以这段代码的空间复杂度为O(n)</p>
<h5 id="O-n-2-1"><a href="#O-n-2-1" class="headerlink" title="O(n^2)"></a>O(n^2)</h5><p>平方空间复杂度与O(n)类似，new出来的数组如果是n * n的多维数组，那就是O(n^2)复杂度，如果是m * n的多维数组,那就是O(m * n)复杂度。</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">int[][] a = new int[n][n];</span><br><span class="line">int[][] a = new int[m][n];</span><br></pre></td></tr></table></figure>
<h2 id="总结"><a href="#总结" class="headerlink" title="总结"></a>总结</h2><p>今天我们学习了复杂度分析的相关基础知识，包括了时间复杂度和空间复杂度，主要是用来分析算法执行效率与数据规模之间的增长关系。</p>
<p>常见的复杂度，从低阶到高阶有：O(1)、O(logn)、O(n)、O(nlogn)、O(n^2)，几乎包括了后面学习的所有数据结构和算法的复杂度，希望大家可以好好的掌握。</p>
<p><img src="https://tva1.sinaimg.cn/large/0082zybply1gc7l65ikduj30vq0hs3zg.jpg" alt></p>
<p>时间复杂度分析还有最好，最坏，平均，均摊时间复杂度等话题需要好好讨论，请关注后续文章，我们会在刷题过程中穿插讲解。</p>
<h4 id="参考资料"><a href="#参考资料" class="headerlink" title="参考资料"></a>参考资料</h4><ul>
<li>analysis_of_algorithms: <a href="https://en.wikipedia.org/wiki/Master_theorem_(analysis_of_algorithms)" target="_blank" rel="noopener">https://en.wikipedia.org/wiki/Master_theorem_(analysis_of_algorithms)</a></li>
</ul>
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