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<span class="md-ellipsis">
第 2 章 &nbsp; 复杂度
第 2 章 &nbsp; 时空复杂度
</span>
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第 2 章 &nbsp; 复杂度
第 2 章 &nbsp; 时空复杂度
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第 6 章 &nbsp; 散列表
第 6 章 &nbsp; 哈希表
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第 6 章 &nbsp; 散列表
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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>
<div class="arithmatex">\[
T(h) = 2^0h + 2^1(h-1) + 2^2(h-2) + \cdots + 2^{(h-1)}\times1
T(h) = 2^0h + 2^1(h-1) + 2^2(h-2) + \dots + 2^{(h-1)}\times1
\]</div>
<p>化简上式需要借助中学的数列知识，先对 <span class="arithmatex">\(T(h)\)</span> 乘以 <span class="arithmatex">\(2\)</span> ，得到</p>
<div class="arithmatex">\[
\begin{aligned}
T(h) &amp; = 2^0h + 2^1(h-1) + 2^2(h-2) + \cdots + 2^{h-1}\times1 \newline
2 T(h) &amp; = 2^1h + 2^2(h-1) + 2^3(h-2) + \cdots + 2^{h}\times1 \newline
T(h) &amp; = 2^0h + 2^1(h-1) + 2^2(h-2) + \dots + 2^{h-1}\times1 \newline
2 T(h) &amp; = 2^1h + 2^2(h-1) + 2^3(h-2) + \dots + 2^{h}\times1 \newline
\end{aligned}
\]</div>
<p>使用错位相减法，用下式 <span class="arithmatex">\(2 T(h)\)</span> 减去上式 <span class="arithmatex">\(T(h)\)</span> ，可得</p>
<div class="arithmatex">\[
2T(h) - T(h) = T(h) = -2^0h + 2^1 + 2^2 + \cdots + 2^{h-1} + 2^h
2T(h) - T(h) = T(h) = -2^0h + 2^1 + 2^2 + \dots + 2^{h-1} + 2^h
\]</div>
<p>观察上式，发现 <span class="arithmatex">\(T(h)\)</span> 是一个等比数列，可直接使用求和公式，得到时间复杂度为</p>
<div class="arithmatex">\[