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krahets
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<p>设元素数量为 <span class="arithmatex">\(n\)</span> ,则最后一个元素入堆的时间复杂度为 <span class="arithmatex">\(O(\log n)\)</span> 。在依次添加元素时,堆的平均长度为 <span class="arithmatex">\(\frac{n}{2}\)</span> ,因此该方法的总体时间复杂度为 <span class="arithmatex">\(O(n \log n)\)</span></p>
<h2 id="822">8.2.2. &nbsp; 基于堆化操作实现<a class="headerlink" href="#822" title="Permanent link">&para;</a></h2>
<p>有趣的是,存在一种更高效的建堆方法,其时间复杂度仅为 <span class="arithmatex">\(O(n)\)</span> 。我们先将列表所有元素原封不动添加到堆中,<strong>然后迭代地对各个节点执行“从顶至底堆化”</strong>。当然,<strong>我们不需要对叶节点执行堆化操作</strong>,因为它们没有子节点。</p>
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<div class="highlight"><span class="filename">my_heap.java</span><pre><span></span><code><a id="__codelineno-0-1" name="__codelineno-0-1" href="#__codelineno-0-1"></a><span class="cm">/* 构造方法,根据输入列表建堆 */</span>
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<a id="__codelineno-9-12" name="__codelineno-9-12" href="#__codelineno-9-12"></a><span class="p">}</span>
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<div class="highlight"><span class="filename">my_heap.dart</span><pre><span></span><code><a id="__codelineno-10-1" name="__codelineno-10-1" href="#__codelineno-10-1"></a><span class="cm">/* 构造方法,根据输入列表建堆 */</span>
<a id="__codelineno-10-2" name="__codelineno-10-2" href="#__codelineno-10-2"></a><span class="n">MaxHeap</span><span class="p">(</span><span class="n">List</span><span class="o">&lt;</span><span class="kt">int</span><span class="o">&gt;</span><span class="w"> </span><span class="n">nums</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
<a id="__codelineno-10-3" name="__codelineno-10-3" href="#__codelineno-10-3"></a><span class="w"> </span><span class="c1">// 将列表元素原封不动添加进堆</span>
<a id="__codelineno-10-4" name="__codelineno-10-4" href="#__codelineno-10-4"></a><span class="w"> </span><span class="n">_maxHeap</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">nums</span><span class="p">;</span>
<a id="__codelineno-10-5" name="__codelineno-10-5" href="#__codelineno-10-5"></a><span class="w"> </span><span class="c1">// 堆化除叶节点以外的其他所有节点</span>
<a id="__codelineno-10-6" name="__codelineno-10-6" href="#__codelineno-10-6"></a><span class="w"> </span><span class="k">for</span><span class="w"> </span><span class="p">(</span><span class="kt">int</span><span class="w"> </span><span class="n">i</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">_parent</span><span class="p">(</span><span class="n">size</span><span class="p">()</span><span class="w"> </span><span class="o">-</span><span class="w"> </span><span class="m">1</span><span class="p">);</span><span class="w"> </span><span class="n">i</span><span class="w"> </span><span class="o">&gt;=</span><span class="w"> </span><span class="m">0</span><span class="p">;</span><span class="w"> </span><span class="n">i</span><span class="o">--</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
<a id="__codelineno-10-7" name="__codelineno-10-7" href="#__codelineno-10-7"></a><span class="w"> </span><span class="n">_siftDown</span><span class="p">(</span><span class="n">i</span><span class="p">);</span>
<a id="__codelineno-10-8" name="__codelineno-10-8" href="#__codelineno-10-8"></a><span class="w"> </span><span class="p">}</span>
<a id="__codelineno-10-9" name="__codelineno-10-9" href="#__codelineno-10-9"></a><span class="p">}</span>
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<h2 id="823">8.2.3. &nbsp; 复杂度分析<a class="headerlink" href="#823" title="Permanent link">&para;</a></h2>

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