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<div class="md-content" data-md-component="content">
<article class="md-content__inner md-typeset">
<h1 id="_1">链表、栈和队列<a class="headerlink" href="#_1" title="Permanent link">&para;</a></h1>
<p><em>链表、栈和队列是更复杂数据结构的构建模块。本文件涵盖它们的运行机制，然后构建关键模式：快慢指针、单调栈和基于堆的优先队列，通过逐步增加难度的题目，并在每一步指出常见陷阱。</em></p>
<ul>
<li>数组提供了快速的随机访问但插入代价高。<strong>链表</strong>提供了快速插入但没有随机访问。<strong>栈</strong>和<strong>队列</strong>将访问限制在一端或两端，而正是这种限制使它们强大：通过限制你能做的事情，它们简化了你需要考虑的事情。</li>
</ul>
<h2 id="_2">链表<a class="headerlink" href="#_2" title="Permanent link">&para;</a></h2>
<ul>
<li><strong>单向链表</strong>是一个节点链。每个节点存储一个值和一个指向下一个节点的指针。最后一个节点指向 <code>null</code>。</li>
</ul>
<div class="highlight"><pre><span></span><code><a id="__codelineno-0-1" name="__codelineno-0-1" href="#__codelineno-0-1"></a><span class="k">class</span><span class="w"> </span><span class="nc">ListNode</span><span class="p">:</span>
<a id="__codelineno-0-2" name="__codelineno-0-2" href="#__codelineno-0-2"></a> <span class="k">def</span><span class="w"> </span><span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">val</span><span class="o">=</span><span class="mi">0</span><span class="p">,</span> <span class="nb">next</span><span class="o">=</span><span class="kc">None</span><span class="p">):</span>
<a id="__codelineno-0-3" name="__codelineno-0-3" href="#__codelineno-0-3"></a> <span class="bp">self</span><span class="o">.</span><span class="n">val</span> <span class="o">=</span> <span class="n">val</span>
<a id="__codelineno-0-4" name="__codelineno-0-4" href="#__codelineno-0-4"></a> <span class="bp">self</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="nb">next</span>
</code></pre></div>
<ul>
<li>
<p><strong>相对于数组的优势</strong>：在已知位置插入或删除是 <span class="arithmatex">\(O(1)\)</span>（只需重新指向指针）。无需移动元素。</p>
</li>
<li>
<p><strong>劣势</strong>：访问元素 <span class="arithmatex">\(i\)</span> 需要 <span class="arithmatex">\(O(i)\)</span> 次遍历（无随机访问）。缓存局部性差（节点分散在内存中）。</p>
</li>
<li>
<p><strong>双向链表</strong>增加了一个 <code>prev</code> 指针，支持向后遍历。用于 LRU 缓存（常数时间删除任何节点）和浏览器历史（前进/后退）。</p>
</li>
</ul>
<table>
<thead>
<tr>
<th>操作</th>
<th>单向</th>
<th>双向</th>
</tr>
</thead>
<tbody>
<tr>
<td>按索引访问</td>
<td><span class="arithmatex">\(O(n)\)</span></td>
<td><span class="arithmatex">\(O(n)\)</span></td>
</tr>
<tr>
<td>在头部插入</td>
<td><span class="arithmatex">\(O(1)\)</span></td>
<td><span class="arithmatex">\(O(1)\)</span></td>
</tr>
<tr>
<td>在尾部插入</td>
<td><span class="arithmatex">\(O(n)\)</span> 或 <span class="arithmatex">\(O(1)\)</span>*</td>
<td><span class="arithmatex">\(O(1)\)</span></td>
</tr>
<tr>
<td>删除给定节点</td>
<td><span class="arithmatex">\(O(n)\)</span>**</td>
<td><span class="arithmatex">\(O(1)\)</span></td>
</tr>
<tr>
<td>搜索</td>
<td><span class="arithmatex">\(O(n)\)</span></td>
<td><span class="arithmatex">\(O(n)\)</span></td>
</tr>
</tbody>
</table>
<p><em>有尾指针时。</em>*需要前驱节点，需要遍历。</p>
<ul>
<li><strong>哨兵节点</strong>（虚拟头/尾节点）简化了边界情况。没有虚拟头节点时，在头部插入或删除头部需要特殊代码。有了虚拟节点，每个真实节点都有前驱。</li>
</ul>
<div class="highlight"><pre><span></span><code><a id="__codelineno-1-1" name="__codelineno-1-1" href="#__codelineno-1-1"></a><span class="c1"># 无虚拟节点：头部删除需要特殊处理</span>
<a id="__codelineno-1-2" name="__codelineno-1-2" href="#__codelineno-1-2"></a><span class="k">def</span><span class="w"> </span><span class="nf">delete_head</span><span class="p">(</span><span class="n">head</span><span class="p">):</span>
<a id="__codelineno-1-3" name="__codelineno-1-3" href="#__codelineno-1-3"></a> <span class="k">if</span> <span class="ow">not</span> <span class="n">head</span><span class="p">:</span>
<a id="__codelineno-1-4" name="__codelineno-1-4" href="#__codelineno-1-4"></a> <span class="k">return</span> <span class="kc">None</span>
<a id="__codelineno-1-5" name="__codelineno-1-5" href="#__codelineno-1-5"></a> <span class="k">return</span> <span class="n">head</span><span class="o">.</span><span class="n">next</span>
<a id="__codelineno-1-6" name="__codelineno-1-6" href="#__codelineno-1-6"></a>
<a id="__codelineno-1-7" name="__codelineno-1-7" href="#__codelineno-1-7"></a><span class="c1"># 有虚拟节点：统一逻辑</span>
<a id="__codelineno-1-8" name="__codelineno-1-8" href="#__codelineno-1-8"></a><span class="n">dummy</span> <span class="o">=</span> <span class="n">ListNode</span><span class="p">(</span><span class="mi">0</span><span class="p">)</span>
<a id="__codelineno-1-9" name="__codelineno-1-9" href="#__codelineno-1-9"></a><span class="n">dummy</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="n">head</span>
<a id="__codelineno-1-10" name="__codelineno-1-10" href="#__codelineno-1-10"></a><span class="c1"># 现在每次删除都是：prev.next = prev.next.next</span>
</code></pre></div>
<ul>
<li><strong>陷阱</strong>：忘记处理空列表（<code>head is None</code>）或单元素列表。始终测试这些边界情况。</li>
</ul>
<hr />
<h2 id="_3">模式：快慢指针（弗洛伊德算法）<a class="headerlink" href="#_3" title="Permanent link">&para;</a></h2>
<ul>
<li>使用两个以不同速度移动的指针来检测链表的属性。<strong>慢</strong>指针一次移动一步；<strong>快</strong>指针一次移动两步。</li>
</ul>
<h3 id="_4">简单：环形链表<a class="headerlink" href="#_4" title="Permanent link">&para;</a></h3>
<ul>
<li>
<p><strong>问题</strong>：判断一个链表是否有环。</p>
</li>
<li>
<p><strong>模式</strong>：如果有环，快指针最终会追上慢指针（它们会相遇）。如果没有环，快指针会到达 <code>null</code>。</p>
</li>
</ul>
<div class="highlight"><pre><span></span><code><a id="__codelineno-2-1" name="__codelineno-2-1" href="#__codelineno-2-1"></a><span class="k">def</span><span class="w"> </span><span class="nf">has_cycle</span><span class="p">(</span><span class="n">head</span><span class="p">):</span>
<a id="__codelineno-2-2" name="__codelineno-2-2" href="#__codelineno-2-2"></a> <span class="n">slow</span> <span class="o">=</span> <span class="n">fast</span> <span class="o">=</span> <span class="n">head</span>
<a id="__codelineno-2-3" name="__codelineno-2-3" href="#__codelineno-2-3"></a> <span class="k">while</span> <span class="n">fast</span> <span class="ow">and</span> <span class="n">fast</span><span class="o">.</span><span class="n">next</span><span class="p">:</span>
<a id="__codelineno-2-4" name="__codelineno-2-4" href="#__codelineno-2-4"></a> <span class="n">slow</span> <span class="o">=</span> <span class="n">slow</span><span class="o">.</span><span class="n">next</span>
<a id="__codelineno-2-5" name="__codelineno-2-5" href="#__codelineno-2-5"></a> <span class="n">fast</span> <span class="o">=</span> <span class="n">fast</span><span class="o">.</span><span class="n">next</span><span class="o">.</span><span class="n">next</span>
<a id="__codelineno-2-6" name="__codelineno-2-6" href="#__codelineno-2-6"></a> <span class="k">if</span> <span class="n">slow</span> <span class="o">==</span> <span class="n">fast</span><span class="p">:</span>
<a id="__codelineno-2-7" name="__codelineno-2-7" href="#__codelineno-2-7"></a> <span class="k">return</span> <span class="kc">True</span>
<a id="__codelineno-2-8" name="__codelineno-2-8" href="#__codelineno-2-8"></a> <span class="k">return</span> <span class="kc">False</span>
</code></pre></div>
<ul>
<li>
<p><strong>为什么有效</strong>：如果环的长度为 <span class="arithmatex">\(c\)</span>，快指针每步缩小1个节点的差距。它们必在慢指针进入环后的 <span class="arithmatex">\(c\)</span> 步内相遇。</p>
</li>
<li>
<p><strong>陷阱</strong>：检查 <code>fast and fast.next</code>（而不仅仅是 <code>fast.next</code>）。如果 <code>fast</code> 是 <code>None</code>，调用 <code>fast.next</code> 会崩溃。</p>
</li>
</ul>
<h3 id="_5">中等：寻找链表的中间节点<a class="headerlink" href="#_5" title="Permanent link">&para;</a></h3>
<ul>
<li>
<p><strong>问题</strong>：返回中间节点。</p>
</li>
<li>
<p><strong>模式</strong>：当快指针到达末尾时，慢指针在中间。</p>
</li>
</ul>
<div class="highlight"><pre><span></span><code><a id="__codelineno-3-1" name="__codelineno-3-1" href="#__codelineno-3-1"></a><span class="k">def</span><span class="w"> </span><span class="nf">find_middle</span><span class="p">(</span><span class="n">head</span><span class="p">):</span>
<a id="__codelineno-3-2" name="__codelineno-3-2" href="#__codelineno-3-2"></a> <span class="n">slow</span> <span class="o">=</span> <span class="n">fast</span> <span class="o">=</span> <span class="n">head</span>
<a id="__codelineno-3-3" name="__codelineno-3-3" href="#__codelineno-3-3"></a> <span class="k">while</span> <span class="n">fast</span> <span class="ow">and</span> <span class="n">fast</span><span class="o">.</span><span class="n">next</span><span class="p">:</span>
<a id="__codelineno-3-4" name="__codelineno-3-4" href="#__codelineno-3-4"></a> <span class="n">slow</span> <span class="o">=</span> <span class="n">slow</span><span class="o">.</span><span class="n">next</span>
<a id="__codelineno-3-5" name="__codelineno-3-5" href="#__codelineno-3-5"></a> <span class="n">fast</span> <span class="o">=</span> <span class="n">fast</span><span class="o">.</span><span class="n">next</span><span class="o">.</span><span class="n">next</span>
<a id="__codelineno-3-6" name="__codelineno-3-6" href="#__codelineno-3-6"></a> <span class="k">return</span> <span class="n">slow</span> <span class="c1"># slow 在中间（偶数长度时为第二个中间节点）</span>
</code></pre></div>
<h3 id="ii">中等：环形链表 II（寻找环的起点）<a class="headerlink" href="#ii" title="Permanent link">&para;</a></h3>
<ul>
<li>
<p><strong>问题</strong>：返回环开始的节点。</p>
</li>
<li>
<p><strong>模式</strong>：在快指针和慢指针相遇后，将一个指针重置到头部。两者以速度1移动。它们在环的起点相遇。</p>
</li>
</ul>
<div class="highlight"><pre><span></span><code><a id="__codelineno-4-1" name="__codelineno-4-1" href="#__codelineno-4-1"></a><span class="k">def</span><span class="w"> </span><span class="nf">detect_cycle</span><span class="p">(</span><span class="n">head</span><span class="p">):</span>
<a id="__codelineno-4-2" name="__codelineno-4-2" href="#__codelineno-4-2"></a> <span class="n">slow</span> <span class="o">=</span> <span class="n">fast</span> <span class="o">=</span> <span class="n">head</span>
<a id="__codelineno-4-3" name="__codelineno-4-3" href="#__codelineno-4-3"></a> <span class="k">while</span> <span class="n">fast</span> <span class="ow">and</span> <span class="n">fast</span><span class="o">.</span><span class="n">next</span><span class="p">:</span>
<a id="__codelineno-4-4" name="__codelineno-4-4" href="#__codelineno-4-4"></a> <span class="n">slow</span> <span class="o">=</span> <span class="n">slow</span><span class="o">.</span><span class="n">next</span>
<a id="__codelineno-4-5" name="__codelineno-4-5" href="#__codelineno-4-5"></a> <span class="n">fast</span> <span class="o">=</span> <span class="n">fast</span><span class="o">.</span><span class="n">next</span><span class="o">.</span><span class="n">next</span>
<a id="__codelineno-4-6" name="__codelineno-4-6" href="#__codelineno-4-6"></a> <span class="k">if</span> <span class="n">slow</span> <span class="o">==</span> <span class="n">fast</span><span class="p">:</span>
<a id="__codelineno-4-7" name="__codelineno-4-7" href="#__codelineno-4-7"></a> <span class="c1"># 将一个指针重置到头部</span>
<a id="__codelineno-4-8" name="__codelineno-4-8" href="#__codelineno-4-8"></a> <span class="n">slow</span> <span class="o">=</span> <span class="n">head</span>
<a id="__codelineno-4-9" name="__codelineno-4-9" href="#__codelineno-4-9"></a> <span class="k">while</span> <span class="n">slow</span> <span class="o">!=</span> <span class="n">fast</span><span class="p">:</span>
<a id="__codelineno-4-10" name="__codelineno-4-10" href="#__codelineno-4-10"></a> <span class="n">slow</span> <span class="o">=</span> <span class="n">slow</span><span class="o">.</span><span class="n">next</span>
<a id="__codelineno-4-11" name="__codelineno-4-11" href="#__codelineno-4-11"></a> <span class="n">fast</span> <span class="o">=</span> <span class="n">fast</span><span class="o">.</span><span class="n">next</span>
<a id="__codelineno-4-12" name="__codelineno-4-12" href="#__codelineno-4-12"></a> <span class="k">return</span> <span class="n">slow</span>
<a id="__codelineno-4-13" name="__codelineno-4-13" href="#__codelineno-4-13"></a> <span class="k">return</span> <span class="kc">None</span>
</code></pre></div>
<ul>
<li><strong>为什么有效</strong>：设从头到环起点的距离为 <span class="arithmatex">\(a\)</span>，从环起点到相遇点的距离为 <span class="arithmatex">\(b\)</span>。慢指针走了 <span class="arithmatex">\(a + b\)</span>。快指针走了 <span class="arithmatex">\(2(a + b)\)</span>。差值为一整圈：<span class="arithmatex">\(a + b = c\)</span>（环长）。所以 <span class="arithmatex">\(a = c - b\)</span>：从头到环起点的距离等于从相遇点到环起点的距离（沿环向前走）。</li>
</ul>
<h3 id="k">困难：K个一组反转链表<a class="headerlink" href="#k" title="Permanent link">&para;</a></h3>
<ul>
<li><strong>问题</strong>：反转链表中每 <span class="arithmatex">\(k\)</span> 个连续节点。</li>
</ul>
<div class="highlight"><pre><span></span><code><a id="__codelineno-5-1" name="__codelineno-5-1" href="#__codelineno-5-1"></a><span class="k">def</span><span class="w"> </span><span class="nf">reverse_k_group</span><span class="p">(</span><span class="n">head</span><span class="p">,</span> <span class="n">k</span><span class="p">):</span>
<a id="__codelineno-5-2" name="__codelineno-5-2" href="#__codelineno-5-2"></a> <span class="c1"># 检查是否还有 k 个节点</span>
<a id="__codelineno-5-3" name="__codelineno-5-3" href="#__codelineno-5-3"></a> <span class="n">node</span> <span class="o">=</span> <span class="n">head</span>
<a id="__codelineno-5-4" name="__codelineno-5-4" href="#__codelineno-5-4"></a> <span class="k">for</span> <span class="n">_</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">k</span><span class="p">):</span>
<a id="__codelineno-5-5" name="__codelineno-5-5" href="#__codelineno-5-5"></a> <span class="k">if</span> <span class="ow">not</span> <span class="n">node</span><span class="p">:</span>
<a id="__codelineno-5-6" name="__codelineno-5-6" href="#__codelineno-5-6"></a> <span class="k">return</span> <span class="n">head</span>
<a id="__codelineno-5-7" name="__codelineno-5-7" href="#__codelineno-5-7"></a> <span class="n">node</span> <span class="o">=</span> <span class="n">node</span><span class="o">.</span><span class="n">next</span>
<a id="__codelineno-5-8" name="__codelineno-5-8" href="#__codelineno-5-8"></a>
<a id="__codelineno-5-9" name="__codelineno-5-9" href="#__codelineno-5-9"></a> <span class="c1"># 反转 k 个节点</span>
<a id="__codelineno-5-10" name="__codelineno-5-10" href="#__codelineno-5-10"></a> <span class="n">prev</span><span class="p">,</span> <span class="n">curr</span> <span class="o">=</span> <span class="kc">None</span><span class="p">,</span> <span class="n">head</span>
<a id="__codelineno-5-11" name="__codelineno-5-11" href="#__codelineno-5-11"></a> <span class="k">for</span> <span class="n">_</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">k</span><span class="p">):</span>
<a id="__codelineno-5-12" name="__codelineno-5-12" href="#__codelineno-5-12"></a> <span class="n">nxt</span> <span class="o">=</span> <span class="n">curr</span><span class="o">.</span><span class="n">next</span>
<a id="__codelineno-5-13" name="__codelineno-5-13" href="#__codelineno-5-13"></a> <span class="n">curr</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="n">prev</span>
<a id="__codelineno-5-14" name="__codelineno-5-14" href="#__codelineno-5-14"></a> <span class="n">prev</span> <span class="o">=</span> <span class="n">curr</span>
<a id="__codelineno-5-15" name="__codelineno-5-15" href="#__codelineno-5-15"></a> <span class="n">curr</span> <span class="o">=</span> <span class="n">nxt</span>
<a id="__codelineno-5-16" name="__codelineno-5-16" href="#__codelineno-5-16"></a>
<a id="__codelineno-5-17" name="__codelineno-5-17" href="#__codelineno-5-17"></a> <span class="c1"># 当前 head 现在是反转后的组的尾节点</span>
<a id="__codelineno-5-18" name="__codelineno-5-18" href="#__codelineno-5-18"></a> <span class="c1"># 递归处理剩余部分</span>
<a id="__codelineno-5-19" name="__codelineno-5-19" href="#__codelineno-5-19"></a> <span class="n">head</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="n">reverse_k_group</span><span class="p">(</span><span class="n">curr</span><span class="p">,</span> <span class="n">k</span><span class="p">)</span>
<a id="__codelineno-5-20" name="__codelineno-5-20" href="#__codelineno-5-20"></a> <span class="k">return</span> <span class="n">prev</span> <span class="c1"># prev 是这组的新头节点</span>
</code></pre></div>
<ul>
<li><strong>陷阱</strong>：原地反转模式（<code>prev, curr, nxt</code>）值得记住。画出来：每一步，你将 <code>curr.next</code> 指回 <code>prev</code>，然后推进所有三个指针。顺序搞错会破坏链表。</li>
</ul>
<hr />
<h2 id="_6">栈<a class="headerlink" href="#_6" title="Permanent link">&para;</a></h2>
<ul>
<li>
<p><strong>栈</strong>是 LIFO（后进先出）：最近添加的元素最先被移除。想象一堆盘子。</p>
</li>
<li>
<p>操作：<code>push(x)</code> 添加到顶部，<code>pop()</code> 从顶部移除，<code>peek()</code> 查看顶部不移除。全部 <span class="arithmatex">\(O(1)\)</span>。</p>
</li>
<li>
<p>栈是<strong>递归</strong>（调用栈）、<strong>表达式求值</strong>（中缀转后缀）和<strong>撤销操作</strong>（每个操作被入栈，撤销时弹出最后一个）背后的隐式结构。</p>
</li>
</ul>
<h3 id="_7">简单：有效的括号<a class="headerlink" href="#_7" title="Permanent link">&para;</a></h3>
<ul>
<li>
<p><strong>问题</strong>：给定一个由括号 <code>()[]{}</code> 组成的字符串，判断它们是否平衡。</p>
</li>
<li>
<p><strong>模式</strong>：将左括号入栈。当看到右括号时，检查栈顶是否匹配。</p>
</li>
</ul>
<div class="highlight"><pre><span></span><code><a id="__codelineno-6-1" name="__codelineno-6-1" href="#__codelineno-6-1"></a><span class="k">def</span><span class="w"> </span><span class="nf">is_valid</span><span class="p">(</span><span class="n">s</span><span class="p">):</span>
<a id="__codelineno-6-2" name="__codelineno-6-2" href="#__codelineno-6-2"></a> <span class="n">stack</span> <span class="o">=</span> <span class="p">[]</span>
<a id="__codelineno-6-3" name="__codelineno-6-3" href="#__codelineno-6-3"></a> <span class="n">matching</span> <span class="o">=</span> <span class="p">{</span><span class="s1">&#39;)&#39;</span><span class="p">:</span> <span class="s1">&#39;(&#39;</span><span class="p">,</span> <span class="s1">&#39;]&#39;</span><span class="p">:</span> <span class="s1">&#39;[&#39;</span><span class="p">,</span> <span class="s1">&#39;}&#39;</span><span class="p">:</span> <span class="s1">&#39;{&#39;</span><span class="p">}</span>
<a id="__codelineno-6-4" name="__codelineno-6-4" href="#__codelineno-6-4"></a>
<a id="__codelineno-6-5" name="__codelineno-6-5" href="#__codelineno-6-5"></a> <span class="k">for</span> <span class="n">char</span> <span class="ow">in</span> <span class="n">s</span><span class="p">:</span>
<a id="__codelineno-6-6" name="__codelineno-6-6" href="#__codelineno-6-6"></a> <span class="k">if</span> <span class="n">char</span> <span class="ow">in</span> <span class="n">matching</span><span class="p">:</span>
<a id="__codelineno-6-7" name="__codelineno-6-7" href="#__codelineno-6-7"></a> <span class="k">if</span> <span class="ow">not</span> <span class="n">stack</span> <span class="ow">or</span> <span class="n">stack</span><span class="p">[</span><span class="o">-</span><span class="mi">1</span><span class="p">]</span> <span class="o">!=</span> <span class="n">matching</span><span class="p">[</span><span class="n">char</span><span class="p">]:</span>
<a id="__codelineno-6-8" name="__codelineno-6-8" href="#__codelineno-6-8"></a> <span class="k">return</span> <span class="kc">False</span>
<a id="__codelineno-6-9" name="__codelineno-6-9" href="#__codelineno-6-9"></a> <span class="n">stack</span><span class="o">.</span><span class="n">pop</span><span class="p">()</span>
<a id="__codelineno-6-10" name="__codelineno-6-10" href="#__codelineno-6-10"></a> <span class="k">else</span><span class="p">:</span>
<a id="__codelineno-6-11" name="__codelineno-6-11" href="#__codelineno-6-11"></a> <span class="n">stack</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">char</span><span class="p">)</span>
<a id="__codelineno-6-12" name="__codelineno-6-12" href="#__codelineno-6-12"></a>
<a id="__codelineno-6-13" name="__codelineno-6-13" href="#__codelineno-6-13"></a> <span class="k">return</span> <span class="nb">len</span><span class="p">(</span><span class="n">stack</span><span class="p">)</span> <span class="o">==</span> <span class="mi">0</span>
</code></pre></div>
<ul>
<li><strong>陷阱</strong>：忘记最后检查 <code>len(stack) == 0</code>。字符串 "(((" 中没有不匹配的情况，但因为没有闭合的括号，它是无效的。</li>
</ul>
<hr />
<h2 id="_8">模式：单调栈<a class="headerlink" href="#_8" title="Permanent link">&para;</a></h2>
<ul>
<li>
<p><strong>单调栈</strong>维护按排序顺序排列的元素（递增或递减）。当新元素会破坏顺序时，你弹出元素直到顺序恢复。</p>
</li>
<li>
<p><strong>何时使用</strong>：问题要求"对每个元素，找到下一个/上一个更大/更小的元素。"栈的总时间复杂度为 <span class="arithmatex">\(O(n)\)</span>，因为每个元素最多被入栈和出栈一次。</p>
</li>
</ul>
<h3 id="_9">中等：每日温度<a class="headerlink" href="#_9" title="Permanent link">&para;</a></h3>
<ul>
<li>
<p><strong>问题</strong>：给定每日温度，对于每一天，找到需要等多少天才会升温。</p>
</li>
<li>
<p><strong>模式</strong>：使用一个索引栈。当当前温度高于栈顶时，弹出并记录距离。</p>
</li>
</ul>
<div class="highlight"><pre><span></span><code><a id="__codelineno-7-1" name="__codelineno-7-1" href="#__codelineno-7-1"></a><span class="k">def</span><span class="w"> </span><span class="nf">daily_temperatures</span><span class="p">(</span><span class="n">temperatures</span><span class="p">):</span>
<a id="__codelineno-7-2" name="__codelineno-7-2" href="#__codelineno-7-2"></a> <span class="n">n</span> <span class="o">=</span> <span class="nb">len</span><span class="p">(</span><span class="n">temperatures</span><span class="p">)</span>
<a id="__codelineno-7-3" name="__codelineno-7-3" href="#__codelineno-7-3"></a> <span class="n">result</span> <span class="o">=</span> <span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="o">*</span> <span class="n">n</span>
<a id="__codelineno-7-4" name="__codelineno-7-4" href="#__codelineno-7-4"></a> <span class="n">stack</span> <span class="o">=</span> <span class="p">[]</span> <span class="c1"># 索引栈，温度递减</span>
<a id="__codelineno-7-5" name="__codelineno-7-5" href="#__codelineno-7-5"></a>
<a id="__codelineno-7-6" name="__codelineno-7-6" href="#__codelineno-7-6"></a> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
<a id="__codelineno-7-7" name="__codelineno-7-7" href="#__codelineno-7-7"></a> <span class="k">while</span> <span class="n">stack</span> <span class="ow">and</span> <span class="n">temperatures</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="o">&gt;</span> <span class="n">temperatures</span><span class="p">[</span><span class="n">stack</span><span class="p">[</span><span class="o">-</span><span class="mi">1</span><span class="p">]]:</span>
<a id="__codelineno-7-8" name="__codelineno-7-8" href="#__codelineno-7-8"></a> <span class="n">prev</span> <span class="o">=</span> <span class="n">stack</span><span class="o">.</span><span class="n">pop</span><span class="p">()</span>
<a id="__codelineno-7-9" name="__codelineno-7-9" href="#__codelineno-7-9"></a> <span class="n">result</span><span class="p">[</span><span class="n">prev</span><span class="p">]</span> <span class="o">=</span> <span class="n">i</span> <span class="o">-</span> <span class="n">prev</span>
<a id="__codelineno-7-10" name="__codelineno-7-10" href="#__codelineno-7-10"></a> <span class="n">stack</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">i</span><span class="p">)</span>
<a id="__codelineno-7-11" name="__codelineno-7-11" href="#__codelineno-7-11"></a>
<a id="__codelineno-7-12" name="__codelineno-7-12" href="#__codelineno-7-12"></a> <span class="k">return</span> <span class="n">result</span>
</code></pre></div>
<ul>
<li>
<p>每个元素被入栈一次，最多出栈一次：总计 <span class="arithmatex">\(O(n)\)</span>。</p>
</li>
<li>
<p><strong>陷阱</strong>：在栈中存储索引（而非值）。你需要索引来计算距离。</p>
</li>
</ul>
<h3 id="_10">困难：柱状图中最大的矩形<a class="headerlink" href="#_10" title="Permanent link">&para;</a></h3>
<ul>
<li>
<p><strong>问题</strong>：给定一个条形高度数组，找出最大矩形的面积。</p>
</li>
<li>
<p><strong>模式</strong>：对于每个条形，找出它可以向左右延伸多远（即，每侧最近的更短条形）。单调递增栈高效地追踪这个信息。</p>
</li>
</ul>
<div class="highlight"><pre><span></span><code><a id="__codelineno-8-1" name="__codelineno-8-1" href="#__codelineno-8-1"></a><span class="k">def</span><span class="w"> </span><span class="nf">largest_rectangle</span><span class="p">(</span><span class="n">heights</span><span class="p">):</span>
<a id="__codelineno-8-2" name="__codelineno-8-2" href="#__codelineno-8-2"></a> <span class="n">stack</span> <span class="o">=</span> <span class="p">[]</span> <span class="c1"># 索引栈，高度递增</span>
<a id="__codelineno-8-3" name="__codelineno-8-3" href="#__codelineno-8-3"></a> <span class="n">max_area</span> <span class="o">=</span> <span class="mi">0</span>
<a id="__codelineno-8-4" name="__codelineno-8-4" href="#__codelineno-8-4"></a> <span class="n">heights</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="mi">0</span><span class="p">)</span> <span class="c1"># 哨兵，用于最后清空栈</span>
<a id="__codelineno-8-5" name="__codelineno-8-5" href="#__codelineno-8-5"></a>
<a id="__codelineno-8-6" name="__codelineno-8-6" href="#__codelineno-8-6"></a> <span class="k">for</span> <span class="n">i</span><span class="p">,</span> <span class="n">h</span> <span class="ow">in</span> <span class="nb">enumerate</span><span class="p">(</span><span class="n">heights</span><span class="p">):</span>
<a id="__codelineno-8-7" name="__codelineno-8-7" href="#__codelineno-8-7"></a> <span class="n">start</span> <span class="o">=</span> <span class="n">i</span>
<a id="__codelineno-8-8" name="__codelineno-8-8" href="#__codelineno-8-8"></a> <span class="k">while</span> <span class="n">stack</span> <span class="ow">and</span> <span class="n">stack</span><span class="p">[</span><span class="o">-</span><span class="mi">1</span><span class="p">][</span><span class="mi">1</span><span class="p">]</span> <span class="o">&gt;</span> <span class="n">h</span><span class="p">:</span>
<a id="__codelineno-8-9" name="__codelineno-8-9" href="#__codelineno-8-9"></a> <span class="n">idx</span><span class="p">,</span> <span class="n">height</span> <span class="o">=</span> <span class="n">stack</span><span class="o">.</span><span class="n">pop</span><span class="p">()</span>
<a id="__codelineno-8-10" name="__codelineno-8-10" href="#__codelineno-8-10"></a> <span class="n">max_area</span> <span class="o">=</span> <span class="nb">max</span><span class="p">(</span><span class="n">max_area</span><span class="p">,</span> <span class="n">height</span> <span class="o">*</span> <span class="p">(</span><span class="n">i</span> <span class="o">-</span> <span class="n">idx</span><span class="p">))</span>
<a id="__codelineno-8-11" name="__codelineno-8-11" href="#__codelineno-8-11"></a> <span class="n">start</span> <span class="o">=</span> <span class="n">idx</span> <span class="c1"># 当前条形可以延伸到被弹出条形开始的位置</span>
<a id="__codelineno-8-12" name="__codelineno-8-12" href="#__codelineno-8-12"></a> <span class="n">stack</span><span class="o">.</span><span class="n">append</span><span class="p">((</span><span class="n">start</span><span class="p">,</span> <span class="n">h</span><span class="p">))</span>
<a id="__codelineno-8-13" name="__codelineno-8-13" href="#__codelineno-8-13"></a>
<a id="__codelineno-8-14" name="__codelineno-8-14" href="#__codelineno-8-14"></a> <span class="n">heights</span><span class="o">.</span><span class="n">pop</span><span class="p">()</span> <span class="c1"># 移除哨兵</span>
<a id="__codelineno-8-15" name="__codelineno-8-15" href="#__codelineno-8-15"></a> <span class="k">return</span> <span class="n">max_area</span>
</code></pre></div>
<ul>
<li>
<p><strong>陷阱</strong>：<code>start = idx</code> 这行很微妙。当我们弹出一个比当前条形更高的条形时，当前条形可以向后延伸至被弹出条形开始的位置（因为中间的所有条形至少和被弹出条形一样高）。缺少这行会得到错误的面积。</p>
</li>
<li>
<p><strong>陷阱</strong>：哨兵 <code>heights.append(0)</code> 确保栈中所有剩余的条形被处理。没有它，那些右侧从未遇到更短条形的条形会被遗漏。</p>
</li>
</ul>
<hr />
<h2 id="_11">队列<a class="headerlink" href="#_11" title="Permanent link">&para;</a></h2>
<ul>
<li>
<p><strong>队列</strong>是 FIFO（先进先出）：元素从后面添加，从前面移除。想象商店里排队。</p>
</li>
<li>
<p><strong>双端队列</strong>（deque）支持在两端 <span class="arithmatex">\(O(1)\)</span> 插入和删除。Python 的 <code>collections.deque</code> 是标准实现。</p>
</li>
<li>
<p>队列是 <strong>BFS</strong>（广度优先搜索，第14章文件04）、<strong>任务调度</strong>和<strong>消息传递</strong>背后的结构。</p>
</li>
</ul>
<h3 id="_12">简单：用栈实现队列<a class="headerlink" href="#_12" title="Permanent link">&para;</a></h3>
<ul>
<li>
<p><strong>问题</strong>：仅使用两个栈实现一个队列。</p>
</li>
<li>
<p><strong>模式</strong>：使用一个栈进行入队操作，一个栈进行出队操作。当出队栈为空时，将所有元素从入队栈转移到出队栈（反转顺序）。</p>
</li>
</ul>
<div class="highlight"><pre><span></span><code><a id="__codelineno-9-1" name="__codelineno-9-1" href="#__codelineno-9-1"></a><span class="k">class</span><span class="w"> </span><span class="nc">MyQueue</span><span class="p">:</span>
<a id="__codelineno-9-2" name="__codelineno-9-2" href="#__codelineno-9-2"></a> <span class="k">def</span><span class="w"> </span><span class="fm">__init__</span><span class="p">(</span><span class="bp">self</span><span class="p">):</span>
<a id="__codelineno-9-3" name="__codelineno-9-3" href="#__codelineno-9-3"></a> <span class="bp">self</span><span class="o">.</span><span class="n">push_stack</span> <span class="o">=</span> <span class="p">[]</span>
<a id="__codelineno-9-4" name="__codelineno-9-4" href="#__codelineno-9-4"></a> <span class="bp">self</span><span class="o">.</span><span class="n">pop_stack</span> <span class="o">=</span> <span class="p">[]</span>
<a id="__codelineno-9-5" name="__codelineno-9-5" href="#__codelineno-9-5"></a>
<a id="__codelineno-9-6" name="__codelineno-9-6" href="#__codelineno-9-6"></a> <span class="k">def</span><span class="w"> </span><span class="nf">push</span><span class="p">(</span><span class="bp">self</span><span class="p">,</span> <span class="n">x</span><span class="p">):</span>
<a id="__codelineno-9-7" name="__codelineno-9-7" href="#__codelineno-9-7"></a> <span class="bp">self</span><span class="o">.</span><span class="n">push_stack</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">x</span><span class="p">)</span>
<a id="__codelineno-9-8" name="__codelineno-9-8" href="#__codelineno-9-8"></a>
<a id="__codelineno-9-9" name="__codelineno-9-9" href="#__codelineno-9-9"></a> <span class="k">def</span><span class="w"> </span><span class="nf">pop</span><span class="p">(</span><span class="bp">self</span><span class="p">):</span>
<a id="__codelineno-9-10" name="__codelineno-9-10" href="#__codelineno-9-10"></a> <span class="k">if</span> <span class="ow">not</span> <span class="bp">self</span><span class="o">.</span><span class="n">pop_stack</span><span class="p">:</span>
<a id="__codelineno-9-11" name="__codelineno-9-11" href="#__codelineno-9-11"></a> <span class="k">while</span> <span class="bp">self</span><span class="o">.</span><span class="n">push_stack</span><span class="p">:</span>
<a id="__codelineno-9-12" name="__codelineno-9-12" href="#__codelineno-9-12"></a> <span class="bp">self</span><span class="o">.</span><span class="n">pop_stack</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="bp">self</span><span class="o">.</span><span class="n">push_stack</span><span class="o">.</span><span class="n">pop</span><span class="p">())</span>
<a id="__codelineno-9-13" name="__codelineno-9-13" href="#__codelineno-9-13"></a> <span class="k">return</span> <span class="bp">self</span><span class="o">.</span><span class="n">pop_stack</span><span class="o">.</span><span class="n">pop</span><span class="p">()</span>
<a id="__codelineno-9-14" name="__codelineno-9-14" href="#__codelineno-9-14"></a>
<a id="__codelineno-9-15" name="__codelineno-9-15" href="#__codelineno-9-15"></a> <span class="k">def</span><span class="w"> </span><span class="nf">peek</span><span class="p">(</span><span class="bp">self</span><span class="p">):</span>
<a id="__codelineno-9-16" name="__codelineno-9-16" href="#__codelineno-9-16"></a> <span class="k">if</span> <span class="ow">not</span> <span class="bp">self</span><span class="o">.</span><span class="n">pop_stack</span><span class="p">:</span>
<a id="__codelineno-9-17" name="__codelineno-9-17" href="#__codelineno-9-17"></a> <span class="k">while</span> <span class="bp">self</span><span class="o">.</span><span class="n">push_stack</span><span class="p">:</span>
<a id="__codelineno-9-18" name="__codelineno-9-18" href="#__codelineno-9-18"></a> <span class="bp">self</span><span class="o">.</span><span class="n">pop_stack</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="bp">self</span><span class="o">.</span><span class="n">push_stack</span><span class="o">.</span><span class="n">pop</span><span class="p">())</span>
<a id="__codelineno-9-19" name="__codelineno-9-19" href="#__codelineno-9-19"></a> <span class="k">return</span> <span class="bp">self</span><span class="o">.</span><span class="n">pop_stack</span><span class="p">[</span><span class="o">-</span><span class="mi">1</span><span class="p">]</span>
<a id="__codelineno-9-20" name="__codelineno-9-20" href="#__codelineno-9-20"></a>
<a id="__codelineno-9-21" name="__codelineno-9-21" href="#__codelineno-9-21"></a> <span class="k">def</span><span class="w"> </span><span class="nf">empty</span><span class="p">(</span><span class="bp">self</span><span class="p">):</span>
<a id="__codelineno-9-22" name="__codelineno-9-22" href="#__codelineno-9-22"></a> <span class="k">return</span> <span class="ow">not</span> <span class="bp">self</span><span class="o">.</span><span class="n">push_stack</span> <span class="ow">and</span> <span class="ow">not</span> <span class="bp">self</span><span class="o">.</span><span class="n">pop_stack</span>
</code></pre></div>
<ul>
<li>每次操作的平摊复杂度 <span class="arithmatex">\(O(1)\)</span>：每个元素最多在两个栈之间移动一次。</li>
</ul>
<hr />
<h2 id="_13">优先队列和堆<a class="headerlink" href="#_13" title="Permanent link">&para;</a></h2>
<ul>
<li>
<p><strong>优先队列</strong>总是返回最小（或最大）的元素，不论插入顺序。标准实现是<strong>二叉堆</strong>。</p>
</li>
<li>
<p><strong>最小堆</strong>是一棵完全二叉树，其中每个父节点都小于其子节点。最小值总是在根节点。以数组形式存储：节点 <span class="arithmatex">\(i\)</span> 的子节点在位置 <span class="arithmatex">\(2i + 1\)</span> 和 <span class="arithmatex">\(2i + 2\)</span>。</p>
</li>
</ul>
<table>
<thead>
<tr>
<th>操作</th>
<th>时间</th>
</tr>
</thead>
<tbody>
<tr>
<td>插入</td>
<td><span class="arithmatex">\(O(\log n)\)</span></td>
</tr>
<tr>
<td>获取最小值</td>
<td><span class="arithmatex">\(O(1)\)</span></td>
</tr>
<tr>
<td>提取最小值</td>
<td><span class="arithmatex">\(O(\log n)\)</span></td>
</tr>
<tr>
<td>从数组构建堆</td>
<td><span class="arithmatex">\(O(n)\)</span></td>
</tr>
</tbody>
</table>
<ul>
<li>Python 的 <code>heapq</code> 模块提供了最小堆。对于最大堆，将值取反。</li>
</ul>
<div class="highlight"><pre><span></span><code><a id="__codelineno-10-1" name="__codelineno-10-1" href="#__codelineno-10-1"></a><span class="kn">import</span><span class="w"> </span><span class="nn">heapq</span>
<a id="__codelineno-10-2" name="__codelineno-10-2" href="#__codelineno-10-2"></a>
<a id="__codelineno-10-3" name="__codelineno-10-3" href="#__codelineno-10-3"></a><span class="c1"># 最小堆</span>
<a id="__codelineno-10-4" name="__codelineno-10-4" href="#__codelineno-10-4"></a><span class="n">h</span> <span class="o">=</span> <span class="p">[]</span>
<a id="__codelineno-10-5" name="__codelineno-10-5" href="#__codelineno-10-5"></a><span class="n">heapq</span><span class="o">.</span><span class="n">heappush</span><span class="p">(</span><span class="n">h</span><span class="p">,</span> <span class="mi">5</span><span class="p">)</span>
<a id="__codelineno-10-6" name="__codelineno-10-6" href="#__codelineno-10-6"></a><span class="n">heapq</span><span class="o">.</span><span class="n">heappush</span><span class="p">(</span><span class="n">h</span><span class="p">,</span> <span class="mi">2</span><span class="p">)</span>
<a id="__codelineno-10-7" name="__codelineno-10-7" href="#__codelineno-10-7"></a><span class="n">heapq</span><span class="o">.</span><span class="n">heappush</span><span class="p">(</span><span class="n">h</span><span class="p">,</span> <span class="mi">8</span><span class="p">)</span>
<a id="__codelineno-10-8" name="__codelineno-10-8" href="#__codelineno-10-8"></a><span class="nb">print</span><span class="p">(</span><span class="n">heapq</span><span class="o">.</span><span class="n">heappop</span><span class="p">(</span><span class="n">h</span><span class="p">))</span> <span class="c1"># 2（最小）</span>
<a id="__codelineno-10-9" name="__codelineno-10-9" href="#__codelineno-10-9"></a>
<a id="__codelineno-10-10" name="__codelineno-10-10" href="#__codelineno-10-10"></a><span class="c1"># 最大堆技巧：取反</span>
<a id="__codelineno-10-11" name="__codelineno-10-11" href="#__codelineno-10-11"></a><span class="n">heapq</span><span class="o">.</span><span class="n">heappush</span><span class="p">(</span><span class="n">h</span><span class="p">,</span> <span class="o">-</span><span class="mi">10</span><span class="p">)</span>
<a id="__codelineno-10-12" name="__codelineno-10-12" href="#__codelineno-10-12"></a><span class="nb">print</span><span class="p">(</span><span class="o">-</span><span class="n">heapq</span><span class="o">.</span><span class="n">heappop</span><span class="p">(</span><span class="n">h</span><span class="p">))</span> <span class="c1"># 10（最大）</span>
</code></pre></div>
<h3 id="k_1">中等：数组中的第 K 个最大元素<a class="headerlink" href="#k_1" title="Permanent link">&para;</a></h3>
<ul>
<li>
<p><strong>问题</strong>：找到第 k 个最大的元素。</p>
</li>
<li>
<p><strong>模式</strong>：维护一个大小为 <span class="arithmatex">\(k\)</span> 的最小堆。堆的根节点就是第 k 大的元素。如果堆有 <span class="arithmatex">\(k\)</span> 个元素且新元素大于根节点，则替换根节点。</p>
</li>
</ul>
<div class="highlight"><pre><span></span><code><a id="__codelineno-11-1" name="__codelineno-11-1" href="#__codelineno-11-1"></a><span class="kn">import</span><span class="w"> </span><span class="nn">heapq</span>
<a id="__codelineno-11-2" name="__codelineno-11-2" href="#__codelineno-11-2"></a>
<a id="__codelineno-11-3" name="__codelineno-11-3" href="#__codelineno-11-3"></a><span class="k">def</span><span class="w"> </span><span class="nf">find_kth_largest</span><span class="p">(</span><span class="n">nums</span><span class="p">,</span> <span class="n">k</span><span class="p">):</span>
<a id="__codelineno-11-4" name="__codelineno-11-4" href="#__codelineno-11-4"></a> <span class="n">heap</span> <span class="o">=</span> <span class="n">nums</span><span class="p">[:</span><span class="n">k</span><span class="p">]</span>
<a id="__codelineno-11-5" name="__codelineno-11-5" href="#__codelineno-11-5"></a> <span class="n">heapq</span><span class="o">.</span><span class="n">heapify</span><span class="p">(</span><span class="n">heap</span><span class="p">)</span> <span class="c1"># O(k)</span>
<a id="__codelineno-11-6" name="__codelineno-11-6" href="#__codelineno-11-6"></a>
<a id="__codelineno-11-7" name="__codelineno-11-7" href="#__codelineno-11-7"></a> <span class="k">for</span> <span class="n">num</span> <span class="ow">in</span> <span class="n">nums</span><span class="p">[</span><span class="n">k</span><span class="p">:]:</span>
<a id="__codelineno-11-8" name="__codelineno-11-8" href="#__codelineno-11-8"></a> <span class="k">if</span> <span class="n">num</span> <span class="o">&gt;</span> <span class="n">heap</span><span class="p">[</span><span class="mi">0</span><span class="p">]:</span>
<a id="__codelineno-11-9" name="__codelineno-11-9" href="#__codelineno-11-9"></a> <span class="n">heapq</span><span class="o">.</span><span class="n">heapreplace</span><span class="p">(</span><span class="n">heap</span><span class="p">,</span> <span class="n">num</span><span class="p">)</span> <span class="c1"># 弹出最小值，推入 num：O(log k)</span>
<a id="__codelineno-11-10" name="__codelineno-11-10" href="#__codelineno-11-10"></a>
<a id="__codelineno-11-11" name="__codelineno-11-11" href="#__codelineno-11-11"></a> <span class="k">return</span> <span class="n">heap</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span>
</code></pre></div>
<ul>
<li>
<p><span class="arithmatex">\(O(n \log k)\)</span> 时间，<span class="arithmatex">\(O(k)\)</span> 空间。当 <span class="arithmatex">\(k \ll n\)</span> 时，这比排序（<span class="arithmatex">\(O(n \log n)\)</span>）好得多。</p>
</li>
<li>
<p><strong>陷阱</strong>：使用大小为 <span class="arithmatex">\(n\)</span> 的最大堆并弹出 <span class="arithmatex">\(k\)</span> 次也可行但较慢：<span class="arithmatex">\(O(n + k \log n)\)</span>。大小为 <span class="arithmatex">\(k\)</span> 的最小堆是最优方法。</p>
</li>
</ul>
<h3 id="k_2">困难：合并 K 个排序链表<a class="headerlink" href="#k_2" title="Permanent link">&para;</a></h3>
<ul>
<li>
<p><strong>问题</strong>：合并 <span class="arithmatex">\(k\)</span> 个已排序链表为一个排序链表。</p>
</li>
<li>
<p><strong>模式</strong>：使用一个包含每个链表头节点的最小堆。弹出最小的节点，将其添加到结果中，并将其下一个节点推入堆中。</p>
</li>
</ul>
<div class="highlight"><pre><span></span><code><a id="__codelineno-12-1" name="__codelineno-12-1" href="#__codelineno-12-1"></a><span class="kn">import</span><span class="w"> </span><span class="nn">heapq</span>
<a id="__codelineno-12-2" name="__codelineno-12-2" href="#__codelineno-12-2"></a>
<a id="__codelineno-12-3" name="__codelineno-12-3" href="#__codelineno-12-3"></a><span class="k">def</span><span class="w"> </span><span class="nf">merge_k_lists</span><span class="p">(</span><span class="n">lists</span><span class="p">):</span>
<a id="__codelineno-12-4" name="__codelineno-12-4" href="#__codelineno-12-4"></a> <span class="n">heap</span> <span class="o">=</span> <span class="p">[]</span>
<a id="__codelineno-12-5" name="__codelineno-12-5" href="#__codelineno-12-5"></a> <span class="k">for</span> <span class="n">i</span><span class="p">,</span> <span class="n">lst</span> <span class="ow">in</span> <span class="nb">enumerate</span><span class="p">(</span><span class="n">lists</span><span class="p">):</span>
<a id="__codelineno-12-6" name="__codelineno-12-6" href="#__codelineno-12-6"></a> <span class="k">if</span> <span class="n">lst</span><span class="p">:</span>
<a id="__codelineno-12-7" name="__codelineno-12-7" href="#__codelineno-12-7"></a> <span class="n">heapq</span><span class="o">.</span><span class="n">heappush</span><span class="p">(</span><span class="n">heap</span><span class="p">,</span> <span class="p">(</span><span class="n">lst</span><span class="o">.</span><span class="n">val</span><span class="p">,</span> <span class="n">i</span><span class="p">,</span> <span class="n">lst</span><span class="p">))</span>
<a id="__codelineno-12-8" name="__codelineno-12-8" href="#__codelineno-12-8"></a>
<a id="__codelineno-12-9" name="__codelineno-12-9" href="#__codelineno-12-9"></a> <span class="n">dummy</span> <span class="o">=</span> <span class="n">ListNode</span><span class="p">(</span><span class="mi">0</span><span class="p">)</span>
<a id="__codelineno-12-10" name="__codelineno-12-10" href="#__codelineno-12-10"></a> <span class="n">curr</span> <span class="o">=</span> <span class="n">dummy</span>
<a id="__codelineno-12-11" name="__codelineno-12-11" href="#__codelineno-12-11"></a>
<a id="__codelineno-12-12" name="__codelineno-12-12" href="#__codelineno-12-12"></a> <span class="k">while</span> <span class="n">heap</span><span class="p">:</span>
<a id="__codelineno-12-13" name="__codelineno-12-13" href="#__codelineno-12-13"></a> <span class="n">val</span><span class="p">,</span> <span class="n">i</span><span class="p">,</span> <span class="n">node</span> <span class="o">=</span> <span class="n">heapq</span><span class="o">.</span><span class="n">heappop</span><span class="p">(</span><span class="n">heap</span><span class="p">)</span>
<a id="__codelineno-12-14" name="__codelineno-12-14" href="#__codelineno-12-14"></a> <span class="n">curr</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="n">node</span>
<a id="__codelineno-12-15" name="__codelineno-12-15" href="#__codelineno-12-15"></a> <span class="n">curr</span> <span class="o">=</span> <span class="n">curr</span><span class="o">.</span><span class="n">next</span>
<a id="__codelineno-12-16" name="__codelineno-12-16" href="#__codelineno-12-16"></a> <span class="k">if</span> <span class="n">node</span><span class="o">.</span><span class="n">next</span><span class="p">:</span>
<a id="__codelineno-12-17" name="__codelineno-12-17" href="#__codelineno-12-17"></a> <span class="n">heapq</span><span class="o">.</span><span class="n">heappush</span><span class="p">(</span><span class="n">heap</span><span class="p">,</span> <span class="p">(</span><span class="n">node</span><span class="o">.</span><span class="n">next</span><span class="o">.</span><span class="n">val</span><span class="p">,</span> <span class="n">i</span><span class="p">,</span> <span class="n">node</span><span class="o">.</span><span class="n">next</span><span class="p">))</span>
<a id="__codelineno-12-18" name="__codelineno-12-18" href="#__codelineno-12-18"></a>
<a id="__codelineno-12-19" name="__codelineno-12-19" href="#__codelineno-12-19"></a> <span class="k">return</span> <span class="n">dummy</span><span class="o">.</span><span class="n">next</span>
</code></pre></div>
<ul>
<li>
<p><span class="arithmatex">\(O(n \log k)\)</span>，其中 <span class="arithmatex">\(n\)</span> 是总节点数。堆中最多有 <span class="arithmatex">\(k\)</span> 个元素。</p>
</li>
<li>
<p><strong>陷阱</strong>：堆元组中的 <code>i</code>（索引）是用于打破平局的。没有它，当值相等时 Python 会尝试比较 <code>ListNode</code> 对象，这会崩溃因为 <code>ListNode</code> 不支持 <code>&lt;</code>。索确保了一有效的比较。</p>
</li>
</ul>
<hr />
<h2 id="_14">常见陷阱总结<a class="headerlink" href="#_14" title="Permanent link">&para;</a></h2>
<table>
<thead>
<tr>
<th>陷阱</th>
<th>示例</th>
<th>修复</th>
</tr>
</thead>
<tbody>
<tr>
<td><code>fast.next</code> 上的空指针</td>
<td>循环检测中使用 <code>while fast.next</code></td>
<td>检查 <code>fast and fast.next</code></td>
</tr>
<tr>
<td>未处理空链表</td>
<td>反转 <code>None</code></td>
<td>添加 <code>if not head</code> 守卫</td>
</tr>
<tr>
<td>栈下溢</td>
<td>从空栈弹出</td>
<td>检查 <code>len(stack) &gt; 0</code> 或 <code>if stack</code></td>
</tr>
<tr>
<td>忘记哨兵</td>
<td>直方图遗漏了最后的条形</td>
<td>追加 0 来清空栈</td>
</tr>
<tr>
<td>堆中缺少平局打破</td>
<td>比较不可比较的对象</td>
<td>向堆元组添加索引</td>
</tr>
<tr>
<td>遍历时修改链表</td>
<td>遍历时删除节点</td>
<td>使用 prev/curr 模式或虚拟头节点</td>
</tr>
</tbody>
</table>
<hr />
<h2 id="neetcode">课后练习题（NeetCode）<a class="headerlink" href="#neetcode" title="Permanent link">&para;</a></h2>
<h3 id="_15">链表<a class="headerlink" href="#_15" title="Permanent link">&para;</a></h3>
<ul>
<li><a href="https://neetcode.io/problems/reverse-a-linked-list">反转链表</a> — 基础的原地反转</li>
<li><a href="https://neetcode.io/problems/merge-two-sorted-linked-lists">合并两个有序链表</a> — 双指针合并</li>
<li><a href="https://neetcode.io/problems/linked-list-cycle-detection">环形链表</a> — 快慢指针</li>
<li><a href="https://neetcode.io/problems/reorder-linked-list">重排链表</a> — 找中间 + 反转 + 合并</li>
<li><a href="https://neetcode.io/problems/remove-node-from-end-of-linked-list">删除链表的倒数第 N 个节点</a> — 间距为 <span class="arithmatex">\(n\)</span> 的双指针</li>
<li><a href="https://neetcode.io/problems/lru-cache">LRU 缓存</a> — 哈希表 + 双向链表</li>
</ul>
<h3 id="_16">栈<a class="headerlink" href="#_16" title="Permanent link">&para;</a></h3>
<ul>
<li><a href="https://neetcode.io/problems/validate-parentheses">有效的括号</a> — 括号匹配</li>
<li><a href="https://neetcode.io/problems/minimum-stack">最小栈</a> — 在每层跟踪最小值</li>
<li><a href="https://neetcode.io/problems/evaluate-reverse-polish-notation">逆波兰表达式求值</a> — 基于栈的求值</li>
<li><a href="https://neetcode.io/problems/daily-temperatures">每日温度</a> — 单调递减栈</li>
<li><a href="https://neetcode.io/problems/largest-rectangle-in-histogram">柱状图中最大的矩形</a> — 单调递增栈</li>
<li><a href="https://neetcode.io/problems/car-fleet">车队</a> — 带到达时间的栈</li>
</ul>
<h3 id="_17">堆 / 优先队列<a class="headerlink" href="#_17" title="Permanent link">&para;</a></h3>
<ul>
<li><a href="https://neetcode.io/problems/kth-largest-integer-in-a-stream">数据流中的第 K 大元素</a> — 大小为 <span class="arithmatex">\(k\)</span> 的最小堆</li>
<li><a href="https://neetcode.io/problems/last-stone-weight">最后一块石头的重量</a> — 最大堆模拟</li>
<li><a href="https://neetcode.io/problems/k-closest-points-to-origin">最接近原点的 K 个点</a> — 按距离排序的最小堆</li>
<li><a href="https://neetcode.io/problems/task-scheduler">任务调度器</a> — 贪心 + 最大堆 + 冷却时间</li>
<li><a href="https://neetcode.io/problems/find-median-in-a-data-stream">数据流的中位数</a> — 双堆（下半部分用最大堆，上半部分用最小堆）</li>
</ul>
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