This commit is contained in:
krahets
2023-07-26 08:59:03 +08:00
parent fd34c845bc
commit 974fea7de4
48 changed files with 299 additions and 299 deletions
@@ -5974,8 +5974,8 @@ n! = n \times (n - 1) \times (n - 2) \times \cdots \times 2 \times 1
<h2 id="226">2.2.6. &nbsp; 最差、最佳、平均时间复杂度<a class="headerlink" href="#226" title="Permanent link">&para;</a></h2>
<p><strong>某些算法的时间复杂度不是固定的,而是与输入数据的分布有关</strong>。例如,假设输入一个长度为 <span class="arithmatex">\(n\)</span> 的数组 <code>nums</code> ,其中 <code>nums</code> 由从 <span class="arithmatex">\(1\)</span><span class="arithmatex">\(n\)</span> 的数字组成,但元素顺序是随机打乱的;算法的任务是返回元素 <span class="arithmatex">\(1\)</span> 的索引。我们可以得出以下结论:</p>
<ul>
<li><code>nums = [?, ?, ..., 1]</code> ,即当末尾元素是 <span class="arithmatex">\(1\)</span> 时,需要完整遍历数组,此时达到 <strong>最差时间复杂度 <span class="arithmatex">\(O(n)\)</span></strong></li>
<li><code>nums = [1, ?, ?, ...]</code> ,即当首个数字为 <span class="arithmatex">\(1\)</span> 时,无论数组多长都不需要继续遍历,此时达到 <strong>最佳时间复杂度 <span class="arithmatex">\(\Omega(1)\)</span></strong></li>
<li><code>nums = [?, ?, ..., 1]</code> ,即当末尾元素是 <span class="arithmatex">\(1\)</span> 时,需要完整遍历数组,此时达到 <strong>最差时间复杂度 <span class="arithmatex">\(O(n)\)</span></strong></li>
<li><code>nums = [1, ?, ?, ...]</code> ,即当首个数字为 <span class="arithmatex">\(1\)</span> 时,无论数组多长都不需要继续遍历,此时达到 <strong>最佳时间复杂度 <span class="arithmatex">\(\Omega(1)\)</span></strong></li>
</ul>
<p>“函数渐近上界”使用大 <span class="arithmatex">\(O\)</span> 记号表示,代表「最差时间复杂度」。相应地,“函数渐近下界”用 <span class="arithmatex">\(\Omega\)</span> 记号来表示,代表「最佳时间复杂度」。</p>
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