126 lines
8.6 KiB
HTML
126 lines
8.6 KiB
HTML
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</style><!-- hexo-inject:begin --><!-- hexo-inject:end --></head><body><header><a href="/" class="logo-link"><img src="/logo.png"></a><ul class="nav nav-list"><li class="nav-list-item"><a href="/" target="_self" class="nav-list-link">ALL</a></li><li class="nav-list-item"><a href="/categories/leetcode/" target="_self" class="nav-list-link">LEETCODE</a></li><li class="nav-list-item"><a href="https://bearmiebear.blogspot.com" target="_blank" class="nav-list-link">BEAR</a></li><li class="nav-list-item"><a href="/atom.xml" target="_self" class="nav-list-link">RSS</a></li></ul></header><section class="container"><div class="post"><article class="post-block"><h1 class="post-title">leetcode-169</h1><div class="post-meta"><div class="post-time">2020年3月23日</div></div><div class="post-content"><h3 id="169-__u591A_u6570_u5143_u7D20"><a href="#169-__u591A_u6570_u5143_u7D20" class="headerlink" title="169. 多数元素"></a>169. 多数元素</h3><p><a href="https://leetcode-cn.com/problems/majority-element/" target="_blank" rel="noopener">题目</a></p>
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<!-- hexo-inject:begin --><!-- hexo-inject:end --><a id="more"></a>
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<p>一开始的思路是遍历一遍整个列表，用一个字典去记录每个元素出现的次数，当次数大于 $\cfrac{n}{2}$ 时就可以得出结果。</p>
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<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></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">majorityElement</span><span class="params">(self, nums)</span> -> int:</span></span><br><span class="line"> d = {}</span><br><span class="line"> l = len(nums)</span><br><span class="line"> <span class="keyword">for</span> n <span class="keyword">in</span> nums:</span><br><span class="line"> <span class="keyword">if</span> <span class="keyword">not</span> n <span class="keyword">in</span> d:</span><br><span class="line"> d[n] = <span class="number">0</span></span><br><span class="line"> d[n] = d[n] + <span class="number">1</span></span><br><span class="line"> <span class="keyword">if</span> d[n] > l/<span class="number">2</span>:</span><br><span class="line"> <span class="keyword">return</span> n</span><br><span class="line"><span class="comment"># 64 ms 15.1 MB</span></span><br></pre></td></tr></table></figure>
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<p>Python 也有专门计数的库，写起来更简单一点：</p>
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<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></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">majorityElement</span><span class="params">(self, nums)</span>:</span></span><br><span class="line"> counts = collections.Counter(nums)</span><br><span class="line"> <span class="keyword">return</span> max(counts.keys(), key=counts.get)</span><br><span class="line"><span class="comment"># 44 ms 15.1 MB</span></span><br></pre></td></tr></table></figure>
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<p>由于要找的数出现次数大于 $\cfrac{n}{2}$，脑子里掠过一下<a href="https://zh.wikipedia.org/wiki/%E8%92%99%E5%9C%B0%E5%8D%A1%E7%BE%85%E6%96%B9%E6%B3%95" target="_blank" rel="noopener">蒙特卡罗算法</a>，后来在官方解答中也看到类似的思路了：</p>
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<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></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">majorityElement</span><span class="params">(self, nums)</span>:</span></span><br><span class="line"> majority_count = len(nums)//<span class="number">2</span></span><br><span class="line"> <span class="keyword">while</span> <span class="keyword">True</span>:</span><br><span class="line"> candidate = random.choice(nums)</span><br><span class="line"> <span class="keyword">if</span> sum(<span class="number">1</span> <span class="keyword">for</span> elem <span class="keyword">in</span> nums <span class="keyword">if</span> elem == candidate) > majority_count:</span><br><span class="line"> <span class="keyword">return</span> candidate</span><br><span class="line"></span><br><span class="line"><span class="comment">#作者：LeetCode-Solution</span></span><br><span class="line"><span class="comment">#链接：https://leetcode-cn.com/problems/majority-element/solution/duo-shu-yuan-su-by-leetcode-solution/</span></span><br><span class="line"><span class="comment">#来源：力扣（LeetCode）</span></span><br><span class="line"><span class="comment">#著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。</span></span><br></pre></td></tr></table></figure>
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