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@ -3567,7 +3567,7 @@
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<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>
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<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>
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<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">>=</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>
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<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>
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<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>
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<a id="__codelineno-10-8" name="__codelineno-10-8" href="#__codelineno-10-8"></a><span class="w"> </span><span class="p">}</span>
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<a id="__codelineno-10-9" name="__codelineno-10-9" href="#__codelineno-10-9"></a><span class="p">}</span>
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</code></pre></div>
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@ -3596,7 +3596,7 @@
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<p>将上述两者相乘,可得到建堆过程的时间复杂度为 <span class="arithmatex">\(O(n \log n)\)</span> 。<strong>然而,这个估算结果并不准确,因为我们没有考虑到二叉树底层节点数量远多于顶层节点的特性</strong>。</p>
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<p>接下来我们来进行更为详细的计算。为了减小计算难度,我们假设树是一个“完美二叉树”,该假设不会影响计算结果的正确性。设二叉树(即堆)节点数量为 <span class="arithmatex">\(n\)</span> ,树高度为 <span class="arithmatex">\(h\)</span> 。上文提到,<strong>节点堆化最大迭代次数等于该节点到叶节点的距离,而该距离正是“节点高度”</strong>。</p>
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<p><img alt="完美二叉树的各层节点数量" src="../build_heap.assets/heapify_operations_count.png" /></p>
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<p align="center"> Fig. 完美二叉树的各层节点数量 </p>
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<p align="center"> 图:完美二叉树的各层节点数量 </p>
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<p>因此,我们可以将各层的“节点数量 <span class="arithmatex">\(\times\)</span> 节点高度”求和,<strong>从而得到所有节点的堆化迭代次数的总和</strong>。</p>
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<div class="arithmatex">\[
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