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<p><strong>难度：Easy</strong></p>
<h3 id="题目描述"><a href="#题目描述" class="headerlink" title="题目描述"></a>题目描述</h3><p>每封电子邮件都由一个本地名称和一个域名组成，以@符号分隔。</p>
<p>例如，在alice@leetcode.com中， alice是本地名称，而leetcode.com是域名。</p>
<p>除了小写字母，这些电子邮件还可能包含’,’或’+’。</p>
<p>如果在电子邮件地址的本地名称部分中的某些字符之间添加句点（’.’），则发往那里的邮件将会转发到本地名称中没有点的同一地址。例如，”alice.z@leetcode.com” 和 “alicez@leetcode.com”会转发到同一电子邮件地址。 （请注意，此规则不适用于域名。）<br><a id="more"></a><br>如果在本地名称中添加加号（’+’），则会忽略第一个加号后面的所有内容。这允许过滤某些电子邮件，例如 m.y+name@email.com 将转发到my@email.com。（同样，此规则不适用于域名。）</p>
<p>可以同时使用这两个规则。</p>
<p>给定电子邮件列表emails，我们会向列表中的每个地址发送一封电子邮件。实际收到邮件的不同地址有多少？</p>
<h3 id="示例"><a href="#示例" class="headerlink" title="示例"></a>示例</h3><figure class="highlight shell"><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">输入：["test.email+alex@leetcode.com","test.e.mail+bob.cathy@leetcode.com","testemail+david@lee.tcode.com"]</span><br><span class="line">输出：2</span><br><span class="line">解释：实际收到邮件的是 "testemail@leetcode.com" 和 "testemail@lee.tcode.com"。</span><br></pre></td></tr></table></figure>
<h3 id="提示"><a href="#提示" class="headerlink" title="提示"></a>提示</h3><blockquote>
<ul>
<li>1 <= emails[i].length <= 100</li>
<li>1 <= emails.length <= 100</li>
<li>每封emails[i]都包含有且仅有一个 ‘@’ 字符。</li>
</ul>
</blockquote>
<h3 id="解题思路"><a href="#解题思路" class="headerlink" title="解题思路"></a>解题思路</h3><p>这个题目比较简单，按照题目中的意思来编程即可。首先每封eamil中仅包含一个’@’字符，那么可以以’@’切分为本地名称和域名，域名无需处理，本地名称首先去除’+’后面的，然后再将’.’去掉，放入set中去重，最后返回set的大小。</p>
<p>具体可参加下面的代码。</p>
<h3 id="解题代码"><a href="#解题代码" class="headerlink" title="解题代码"></a>解题代码</h3><p><strong>python3</strong><br><figure class="highlight python"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span>:</span></span><br><span class="line"> <span class="function"><span class="keyword">def</span> <span class="title">numUniqueEmails</span><span class="params">(self, emails)</span>:</span></span><br><span class="line"> <span class="string">"""</span></span><br><span class="line"><span class="string"> :type emails: List[str]</span></span><br><span class="line"><span class="string"> :rtype: int</span></span><br><span class="line"><span class="string"> """</span></span><br><span class="line"> ans = set()</span><br><span class="line"> <span class="keyword">for</span> email <span class="keyword">in</span> emails:</span><br><span class="line"> local_name, domain_name = email.split(<span class="string">'@'</span>)</span><br><span class="line"> idx = local_name.find(<span class="string">'+'</span>)</span><br><span class="line"> <span class="keyword">if</span> idx != <span class="number">-1</span>:</span><br><span class="line"> local_name = local_name[<span class="number">0</span>:idx]</span><br><span class="line"> local_name = local_name.replace(<span class="string">'.'</span>, <span class="string">''</span>)</span><br><span class="line"> ans.add(local_name +<span class="string">'@'</span> + domain_name)</span><br><span class="line"> <span class="keyword">return</span> len(ans)</span><br></pre></td></tr></table></figure></p>
<h3 id="运行时间截图"><a href="#运行时间截图" class="headerlink" title="运行时间截图"></a>运行时间截图</h3><p><img src="https://cutoutsy-blog-1253675385.cos.ap-chengdu.myqcloud.com/blog_leetcode929_runtime.png" alt="runtime"></p>
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