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<p>对于该问题，我们先介绍两种思路比较直接的解法，再介绍效率更高的堆解法。</p>
<h2 id="831">8.3.1 &nbsp; 方法一：遍历选择<a class="headerlink" href="#831" title="Permanent link">&para;</a></h2>
<p>我们可以进行图 8-6 所示的 <span class="arithmatex">\(k\)</span> 轮遍历，分别在每轮中提取第 <span class="arithmatex">\(1\)</span> , <span class="arithmatex">\(2\)</span> , <span class="arithmatex">\(\dots\)</span> , <span class="arithmatex">\(k\)</span> 大的元素，时间复杂度为 <span class="arithmatex">\(O(nk)\)</span> 。</p>
<p>我们可以进行图 8-6 所示的 <span class="arithmatex">\(k\)</span> 轮遍历，分别在每轮中提取第 <span class="arithmatex">\(1\)</span>、<span class="arithmatex">\(2\)</span>、<span class="arithmatex">\(\dots\)</span>、<span class="arithmatex">\(k\)</span> 大的元素，时间复杂度为 <span class="arithmatex">\(O(nk)\)</span> 。</p>
<p>此方法只适用于 <span class="arithmatex">\(k \ll n\)</span> 的情况，因为当 <span class="arithmatex">\(k\)</span> 与 <span class="arithmatex">\(n\)</span> 比较接近时，其时间复杂度趋向于 <span class="arithmatex">\(O(n^2)\)</span> ，非常耗时。</p>
<p><img alt="遍历寻找最大的 k 个元素" src="../top_k.assets/top_k_traversal.png" /></p>
<p align="center"> 图 8-6 &nbsp; 遍历寻找最大的 k 个元素 </p>