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常见陷阱
</span>
</a>
</li>
</ul>
</nav>
</li>
<li class="md-nav__item">
<a href="#_18" class="md-nav__link">
<span class="md-ellipsis">
动态规划
</span>
</a>
<nav class="md-nav" aria-label="动态规划">
<ul class="md-nav__list">
<li class="md-nav__item">
<a href="#_19" class="md-nav__link">
<span class="md-ellipsis">
斐波那契数列的动机
</span>
</a>
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<li class="md-nav__item">
<a href="#dp" class="md-nav__link">
<span class="md-ellipsis">
DP 配方
</span>
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<a href="#_20" class="md-nav__link">
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示例：思路过程
</span>
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<a href="#dp_1" class="md-nav__link">
<span class="md-ellipsis">
如何识别 DP 问题
</span>
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<a href="#dp_2" class="md-nav__link">
<span class="md-ellipsis">
DP 的分类
</span>
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<a href="#vs_1" class="md-nav__link">
<span class="md-ellipsis">
自顶向下 vs 自底向上
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<a href="#_21" class="md-nav__link">
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常见陷阱
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融会贯通
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<div class="md-content" data-md-component="content">
<article class="md-content__inner md-typeset">
<h1 id="o">基础：大O表示法、递归、回溯与动态规划<a class="headerlink" href="#o" title="Permanent link">&para;</a></h1>
<p><em>在深入学习数据结构和算法之前，你需要掌握四个基础概念：衡量效率的大O表示法、将问题分解为子问题的递归、带剪枝的穷举搜索——回溯，以及避免冗余计算的动态规划。本文件从基本原理出发逐一讲解。</em></p>
<ul>
<li>本章后续文件默认你已经熟悉了这四个概念。如果你跳过本文件，那么后面文件中的 <span class="arithmatex">\(O(n \log n)\)</span> 标注、递归树遍历、回溯模板和 DP 状态转移对你来说就会像是魔法而非工程。</li>
</ul>
<h2 id="_1">为什么是模式，而非死记硬背<a class="headerlink" href="#_1" title="Permanent link">&para;</a></h2>
<ul>
<li>
<p>LeetCode、NeetCode 和 HackerRank 上有成千上万的编程题。没有人能记住全部，试图这么做是注定失败的策略。面试官不会从固定题库中选题——他们会修改、组合、伪装。背下来的"两数之和"解法，当面试官问你一个从未见过的变体时毫无用处。</p>
</li>
<li>
<p>好消息是：核心模式大约只有 <strong>15-20 种</strong>（双指针、滑动窗口、BFS/DFS、DP、回溯等）。所有问题，无论表面多新颖，最终都归结为这些模式中的一个或几个组合。面试考的不是你是否见过这道题，而是你是否能<strong>剥离上下文</strong>——故事、具体数据类型、边界情况——识别出底层的模式。</p>
</li>
<li>
<p>考虑这三个问题：</p>
<ul>
<li>"在数组中找到两个数，使其和等于一个目标值。"</li>
<li>"找到两个分子，使其结合能之和等于一个阈值。"</li>
<li>"给定一个账户余额列表，找到两个账户的余额之和等于一笔债务。"</li>
</ul>
</li>
<li>
<p>它们看起来截然不同。但它们是同一个问题：<strong>两数之和</strong>。上下文（数字、分子、账户）无关紧要。其结构是：在集合中搜索补数 → 哈希表查找。</p>
</li>
<li>
<p>这就是本章通过<strong>直觉教授模式</strong>而非通过重复教授解题方法的原因。对于每个模式，我们都会解释：</p>
<ul>
<li><strong>问题中的什么结构特征</strong>指示了这个模式（输入已排序 → 双指针；子数组约束 → 滑动窗口；最优子结构 + 重叠子问题 → DP）。</li>
<li><strong>为什么这个模式有效</strong>——数学或逻辑推理，而不仅仅是"它能给出正确答案"。</li>
<li><strong>如何适配它</strong>——通过展示简单、中等和困难变体，在这些变体中相同的核心思想应用于不同的上下文。</li>
</ul>
</li>
<li>
<p>当你深入理解<em>为什么</em>滑动窗口有效（约束的单调性意味着扩展/收缩就足够了），你就可以将其应用到任何具有该结构的问题上，即使是未曾见过的问题。当你只是背下了"无重复字符的最长子串"的代码，一旦问题发生变化，你就会束手无策。</p>
</li>
<li>
<p>实践策略：</p>
<ol>
<li><strong>学习模式</strong>（本章）。</li>
<li><strong>练习识别模式</strong>，在伪装的问题中（每个文件末尾的 NeetCode 练习题）。</li>
<li><strong>练习实现</strong>，在时间压力下。</li>
<li>面试中：阅读题目 → 剥离上下文 → 识别模式 → 实现。</li>
</ol>
</li>
</ul>
<hr />
<h2 id="o_1">大O表示法<a class="headerlink" href="#o_1" title="Permanent link">&para;</a></h2>
<ul>
<li>
<p>当我们说一个算法"快"或"慢"时，需要一种精确的衡量方式。<strong>大O表示法</strong>描述了随着输入规模 <span class="arithmatex">\(n\)</span> 的增长，算法的运行时间（或空间使用量）如何增长，忽略了常数因子和低阶项。</p>
</li>
<li>
<p>形式化定义：<span class="arithmatex">\(f(n) = O(g(n))\)</span> 意味着存在常数 <span class="arithmatex">\(c &gt; 0\)</span> 和 <span class="arithmatex">\(n_0\)</span>，使得对所有 <span class="arithmatex">\(n \geq n_0\)</span> 有 <span class="arithmatex">\(f(n) \leq c \cdot g(n)\)</span>。通俗地说：对于大规模输入，<span class="arithmatex">\(f\)</span> 的增长速度不超过 <span class="arithmatex">\(g\)</span>。</p>
</li>
<li>
<p>为什么要忽略常数？因为 <span class="arithmatex">\(2n\)</span> 的算法和 <span class="arithmatex">\(5n\)</span> 的算法都是 <span class="arithmatex">\(O(n)\)</span>：它们的扩展方式相同。在更快的计算机上，常数会变，但扩展性不会。大O表示法捕捉了问题的<strong>内在</strong>难度，与硬件无关。</p>
</li>
</ul>
<h3 id="_2">增长率层级<a class="headerlink" href="#_2" title="Permanent link">&para;</a></h3>
<ul>
<li>从最快到最慢：</li>
</ul>
<table>
<thead>
<tr>
<th>大O</th>
<th>名称</th>
<th>示例</th>
<th><span class="arithmatex">\(n = 10^6\)</span> 次操作</th>
</tr>
</thead>
<tbody>
<tr>
<td><span class="arithmatex">\(O(1)\)</span></td>
<td>常数级</td>
<td>数组访问、哈希查找</td>
<td>1</td>
</tr>
<tr>
<td><span class="arithmatex">\(O(\log n)\)</span></td>
<td>对数级</td>
<td>二分查找</td>
<td>20</td>
</tr>
<tr>
<td><span class="arithmatex">\(O(n)\)</span></td>
<td>线性级</td>
<td>线性扫描、单循环</td>
<td><span class="arithmatex">\(10^6\)</span></td>
</tr>
<tr>
<td><span class="arithmatex">\(O(n \log n)\)</span></td>
<td>线性对数级</td>
<td>归并排序、高效排序</td>
<td><span class="arithmatex">\(2 \times 10^7\)</span></td>
</tr>
<tr>
<td><span class="arithmatex">\(O(n^2)\)</span></td>
<td>平方级</td>
<td>嵌套循环、暴力配对</td>
<td><span class="arithmatex">\(10^{12}\)</span>（太慢）</td>
</tr>
<tr>
<td><span class="arithmatex">\(O(n^3)\)</span></td>
<td>立方级</td>
<td>三层嵌套循环、矩阵乘法</td>
<td><span class="arithmatex">\(10^{18}\)</span>（实在太慢）</td>
</tr>
<tr>
<td><span class="arithmatex">\(O(2^n)\)</span></td>
<td>指数级</td>
<td>所有子集、暴力回溯</td>
<td><span class="arithmatex">\(10^{301030}\)</span>（不可能）</td>
</tr>
<tr>
<td><span class="arithmatex">\(O(n!)\)</span></td>
<td>阶乘级</td>
<td>所有排列</td>
<td>荒谬</td>
</tr>
</tbody>
</table>
<ul>
<li>
<p><strong>经验法则</strong>：现代计算机每秒执行约 <span class="arithmatex">\(10^8\)</span>–<span class="arithmatex">\(10^9\)</span> 次简单操作。对于1秒的时间限制：</p>
<ul>
<li><span class="arithmatex">\(O(n)\)</span> 适用于 <span class="arithmatex">\(n \leq 10^8\)</span></li>
<li><span class="arithmatex">\(O(n \log n)\)</span> 适用于 <span class="arithmatex">\(n \leq 10^7\)</span></li>
<li><span class="arithmatex">\(O(n^2)\)</span> 适用于 <span class="arithmatex">\(n \leq 10^4\)</span></li>
<li><span class="arithmatex">\(O(2^n)\)</span> 适用于 <span class="arithmatex">\(n \leq 25\)</span></li>
</ul>
</li>
<li>
<p>这张表能立即告诉你当前方法是否足够快。如果 <span class="arithmatex">\(n = 10^5\)</span> 而你的解法是 <span class="arithmatex">\(O(n^2)\)</span>，那就是 <span class="arithmatex">\(10^{10}\)</span> 次操作——太慢了。你需要一个更好的算法。</p>
</li>
</ul>
<h3 id="o_2">如何分析大O<a class="headerlink" href="#o_2" title="Permanent link">&para;</a></h3>
<ul>
<li><strong>单循环</strong>遍历 <span class="arithmatex">\(n\)</span> 个元素：<span class="arithmatex">\(O(n)\)</span>。</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="n">total</span> <span class="o">=</span> <span class="mi">0</span>
<a id="__codelineno-0-2" name="__codelineno-0-2" href="#__codelineno-0-2"></a><span class="k">for</span> <span class="n">x</span> <span class="ow">in</span> <span class="n">arr</span><span class="p">:</span> <span class="c1"># n 次迭代</span>
<a id="__codelineno-0-3" name="__codelineno-0-3" href="#__codelineno-0-3"></a> <span class="n">total</span> <span class="o">+=</span> <span class="n">x</span> <span class="c1"># 每次迭代 O(1)</span>
<a id="__codelineno-0-4" name="__codelineno-0-4" href="#__codelineno-0-4"></a><span class="c1"># 总计：O(n)</span>
</code></pre></div>
<ul>
<li><strong>嵌套循环</strong>：迭代次数相乘。</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="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> <span class="c1"># n 次迭代</span>
<a id="__codelineno-1-2" name="__codelineno-1-2" href="#__codelineno-1-2"></a> <span class="k">for</span> <span class="n">j</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">n</span><span class="p">):</span> <span class="c1"># 每次 n 次迭代</span>
<a id="__codelineno-1-3" name="__codelineno-1-3" href="#__codelineno-1-3"></a> <span class="n">process</span><span class="p">(</span><span class="n">i</span><span class="p">,</span> <span class="n">j</span><span class="p">)</span> <span class="c1"># O(1)</span>
<a id="__codelineno-1-4" name="__codelineno-1-4" href="#__codelineno-1-4"></a><span class="c1"># 总计：O(n^2)</span>
</code></pre></div>
<ul>
<li><strong>每次减半的循环</strong>：<span class="arithmatex">\(O(\log n)\)</span>。每次迭代将问题规模减半，所以需要 <span class="arithmatex">\(\log_2 n\)</span> 次迭代。</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="n">i</span> <span class="o">=</span> <span class="n">n</span>
<a id="__codelineno-2-2" name="__codelineno-2-2" href="#__codelineno-2-2"></a><span class="k">while</span> <span class="n">i</span> <span class="o">&gt;</span> <span class="mi">0</span><span class="p">:</span>
<a id="__codelineno-2-3" name="__codelineno-2-3" href="#__codelineno-2-3"></a> <span class="n">process</span><span class="p">(</span><span class="n">i</span><span class="p">)</span>
<a id="__codelineno-2-4" name="__codelineno-2-4" href="#__codelineno-2-4"></a> <span class="n">i</span> <span class="o">//=</span> <span class="mi">2</span>
<a id="__codelineno-2-5" name="__codelineno-2-5" href="#__codelineno-2-5"></a><span class="c1"># 总计：O(log n)</span>
</code></pre></div>
<ul>
<li><strong>内循环依赖于外循环的嵌套循环</strong>：</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">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-3-2" name="__codelineno-3-2" href="#__codelineno-3-2"></a> <span class="k">for</span> <span class="n">j</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">i</span><span class="p">):</span> <span class="c1"># j 从 0 到 i-1</span>
<a id="__codelineno-3-3" name="__codelineno-3-3" href="#__codelineno-3-3"></a> <span class="n">process</span><span class="p">(</span><span class="n">i</span><span class="p">,</span> <span class="n">j</span><span class="p">)</span>
<a id="__codelineno-3-4" name="__codelineno-3-4" href="#__codelineno-3-4"></a><span class="c1"># 总计：0 + 1 + 2 + ... + (n-1) = n(n-1)/2 = O(n^2)</span>
</code></pre></div>
<ul>
<li><strong>递归</strong>：写出递推关系并求解（第13章介绍了主定理）。例如，归并排序：<span class="arithmatex">\(T(n) = 2T(n/2) + O(n) = O(n \log n)\)</span>。</li>
</ul>
<h3 id="_3">常见陷阱<a class="headerlink" href="#_3" title="Permanent link">&para;</a></h3>
<ul>
<li><strong>隐藏的循环</strong>：Python 中 <code>x in list</code> 是 <span class="arithmatex">\(O(n)\)</span>（线性扫描），但 <code>x in set</code> 是 <span class="arithmatex">\(O(1)\)</span>。在循环中对列表使用 <code>in</code> 会得到 <span class="arithmatex">\(O(n^2)\)</span>，而不是 <span class="arithmatex">\(O(n)\)</span>。</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="c1"># 不好：O(n^2) — 对列表用 &quot;in&quot; 是 O(n)</span>
<a id="__codelineno-4-2" name="__codelineno-4-2" href="#__codelineno-4-2"></a><span class="k">for</span> <span class="n">x</span> <span class="ow">in</span> <span class="n">arr</span><span class="p">:</span>
<a id="__codelineno-4-3" name="__codelineno-4-3" href="#__codelineno-4-3"></a> <span class="k">if</span> <span class="n">x</span> <span class="ow">in</span> <span class="n">another_list</span><span class="p">:</span>
<a id="__codelineno-4-4" name="__codelineno-4-4" href="#__codelineno-4-4"></a> <span class="n">process</span><span class="p">(</span><span class="n">x</span><span class="p">)</span>
<a id="__codelineno-4-5" name="__codelineno-4-5" href="#__codelineno-4-5"></a>
<a id="__codelineno-4-6" name="__codelineno-4-6" href="#__codelineno-4-6"></a><span class="c1"># 好：O(n) — 先转换为 set</span>
<a id="__codelineno-4-7" name="__codelineno-4-7" href="#__codelineno-4-7"></a><span class="n">another_set</span> <span class="o">=</span> <span class="nb">set</span><span class="p">(</span><span class="n">another_list</span><span class="p">)</span>
<a id="__codelineno-4-8" name="__codelineno-4-8" href="#__codelineno-4-8"></a><span class="k">for</span> <span class="n">x</span> <span class="ow">in</span> <span class="n">arr</span><span class="p">:</span>
<a id="__codelineno-4-9" name="__codelineno-4-9" href="#__codelineno-4-9"></a> <span class="k">if</span> <span class="n">x</span> <span class="ow">in</span> <span class="n">another_set</span><span class="p">:</span>
<a id="__codelineno-4-10" name="__codelineno-4-10" href="#__codelineno-4-10"></a> <span class="n">process</span><span class="p">(</span><span class="n">x</span><span class="p">)</span>
</code></pre></div>
<ul>
<li>
<p><strong>字符串拼接</strong>：Python 中 <code>s += c</code> 每次都会复制整个字符串。在 <span class="arithmatex">\(n\)</span> 次迭代的循环中：<span class="arithmatex">\(O(1 + 2 + \cdots + n) = O(n^2)\)</span>。</p>
</li>
<li>
<p><strong>排序主导</strong>：如果你的算法先排序（<span class="arithmatex">\(O(n \log n)\)</span>）然后做线性扫描（<span class="arithmatex">\(O(n)\)</span>），总复杂度是 <span class="arithmatex">\(O(n \log n)\)</span>——排序占主导。</p>
</li>
<li>
<p><strong>平摊复杂度</strong>：某些操作偶尔很昂贵，但平摊下来很便宜。动态数组的追加操作平摊复杂度为 <span class="arithmatex">\(O(1)\)</span>，因为罕见的 <span class="arithmatex">\(O(n)\)</span> 扩容被分摊到 <span class="arithmatex">\(n\)</span> 次便宜的追加操作中。不要混淆平摊 <span class="arithmatex">\(O(1)\)</span> 和最坏情况 <span class="arithmatex">\(O(1)\)</span>。</p>
</li>
</ul>
<h3 id="_4">空间复杂度<a class="headerlink" href="#_4" title="Permanent link">&para;</a></h3>
<ul>
<li>
<p>空间复杂度遵循同样的大O规则，只是应用于内存使用而非时间。</p>
</li>
<li>
<p><strong>原地</strong>算法使用 <span class="arithmatex">\(O(1)\)</span> 额外空间（不计输入）。快速排序是 <span class="arithmatex">\(O(\log n)\)</span> 空间（递归栈深度）。归并排序是 <span class="arithmatex">\(O(n)\)</span>（合并时使用的临时数组）。</p>
</li>
<li>
<p><strong>递归栈</strong>：每次递归调用都会使用栈空间。深度为 <span class="arithmatex">\(n\)</span> 的递归使用 <span class="arithmatex">\(O(n)\)</span> 空间，即使每次调用没有分配额外内存。这就是为什么在具有 <span class="arithmatex">\(n\)</span> 个节点的图上进行递归 DFS 使用 <span class="arithmatex">\(O(n)\)</span> 空间。</p>
</li>
<li>
<p>面试中，始终同时说明时间和空间复杂度。<span class="arithmatex">\(O(n)\)</span> 时间、<span class="arithmatex">\(O(n)\)</span> 空间的解法通常可以接受，但 <span class="arithmatex">\(O(n)\)</span> 时间、<span class="arithmatex">\(O(1)\)</span> 空间的解法更好。面试官可能会要求你优化其中一个。</p>
</li>
</ul>
<hr />
<h2 id="_5">递归<a class="headerlink" href="#_5" title="Permanent link">&para;</a></h2>
<ul>
<li>
<p><strong>递归</strong>是指函数调用自身来解决同一问题的更小实例。它是处理具有递归结构的问题最自然的方式：树、嵌套数据、分治法和数学序列。</p>
</li>
<li>
<p>每个递归函数都有两部分：</p>
<ol>
<li><strong>基本情况</strong>：可以直接解决的最小的实例（无需递归）。这是递归停止的条件。</li>
<li><strong>递归情况</strong>：将问题分解为更小的子问题，递归求解，然后合并结果。</li>
</ol>
</li>
</ul>
<h3 id="_6">示例：阶乘<a class="headerlink" href="#_6" title="Permanent link">&para;</a></h3>
<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">factorial</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
<a id="__codelineno-5-2" name="__codelineno-5-2" href="#__codelineno-5-2"></a> <span class="k">if</span> <span class="n">n</span> <span class="o">&lt;=</span> <span class="mi">1</span><span class="p">:</span> <span class="c1"># 基本情况</span>
<a id="__codelineno-5-3" name="__codelineno-5-3" href="#__codelineno-5-3"></a> <span class="k">return</span> <span class="mi">1</span>
<a id="__codelineno-5-4" name="__codelineno-5-4" href="#__codelineno-5-4"></a> <span class="k">return</span> <span class="n">n</span> <span class="o">*</span> <span class="n">factorial</span><span class="p">(</span><span class="n">n</span> <span class="o">-</span> <span class="mi">1</span><span class="p">)</span> <span class="c1"># 递归情况</span>
</code></pre></div>
<ul>
<li>
<p><code>factorial(4)</code> 的执行过程：</p>
<ul>
<li><code>factorial(4)</code> 调用 <code>factorial(3)</code></li>
<li><code>factorial(3)</code> 调用 <code>factorial(2)</code></li>
<li><code>factorial(2)</code> 调用 <code>factorial(1)</code></li>
<li><code>factorial(1)</code> 返回 <code>1</code>（基本情况）</li>
<li><code>factorial(2)</code> 返回 <code>2 * 1 = 2</code></li>
<li><code>factorial(3)</code> 返回 <code>3 * 2 = 6</code></li>
<li><code>factorial(4)</code> 返回 <code>4 * 6 = 24</code></li>
</ul>
</li>
<li>
<p>每次调用都被压入<strong>调用栈</strong>。栈一直增长直到到达基本情况，然后随着每次调用的返回而展开。如果递归太深（例如 Python 中的 <code>factorial(1000000)</code>），栈会溢出（<code>RecursionError</code>）。Python 的默认递归限制是 1000。</p>
</li>
</ul>
<h3 id="_7">如何以递归方式思考<a class="headerlink" href="#_7" title="Permanent link">&para;</a></h3>
<ul>
<li>
<p>关键的思维转变是：<strong>信任递归</strong>。在编写递归函数时，假设递归调用已经正确返回了更小子问题的答案。你只需要：</p>
<ol>
<li>处理基本情况。</li>
<li>将问题分解为更小的部分。</li>
<li>合并结果。</li>
</ol>
</li>
<li>
<p>你不需要在脑中跟踪每一次递归调用。这就像试图通过在心里执行每次迭代来理解一个循环。相反，验证："如果递归调用给了我更小输入的正确结果，那么我的组合步骤是否给出了完整输入的正确结果？"</p>
</li>
</ul>
<h3 id="_8">示例：链表上的递归<a class="headerlink" href="#_8" title="Permanent link">&para;</a></h3>
<ul>
<li>递归反转链表：</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">reverse</span><span class="p">(</span><span class="n">head</span><span class="p">):</span>
<a id="__codelineno-6-2" name="__codelineno-6-2" href="#__codelineno-6-2"></a> <span class="k">if</span> <span class="ow">not</span> <span class="n">head</span> <span class="ow">or</span> <span class="ow">not</span> <span class="n">head</span><span class="o">.</span><span class="n">next</span><span class="p">:</span> <span class="c1"># 基本情况：0 或 1 个节点</span>
<a id="__codelineno-6-3" name="__codelineno-6-3" href="#__codelineno-6-3"></a> <span class="k">return</span> <span class="n">head</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="n">new_head</span> <span class="o">=</span> <span class="n">reverse</span><span class="p">(</span><span class="n">head</span><span class="o">.</span><span class="n">next</span><span class="p">)</span> <span class="c1"># 反转剩余部分</span>
<a id="__codelineno-6-6" name="__codelineno-6-6" href="#__codelineno-6-6"></a> <span class="n">head</span><span class="o">.</span><span class="n">next</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="n">head</span> <span class="c1"># 将下一个节点指回当前节点</span>
<a id="__codelineno-6-7" name="__codelineno-6-7" href="#__codelineno-6-7"></a> <span class="n">head</span><span class="o">.</span><span class="n">next</span> <span class="o">=</span> <span class="kc">None</span> <span class="c1"># 当前节点现在成为尾节点</span>
<a id="__codelineno-6-8" name="__codelineno-6-8" href="#__codelineno-6-8"></a> <span class="k">return</span> <span class="n">new_head</span>
</code></pre></div>
<ul>
<li><strong>信任递归</strong>：<code>reverse(head.next)</code> 正确反转了链表的剩余部分并返回新的头节点。我们只需将当前节点附加到末尾。</li>
</ul>
<h3 id="_9">示例：树上的递归<a class="headerlink" href="#_9" title="Permanent link">&para;</a></h3>
<ul>
<li>计算二叉树的高度：</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">height</span><span class="p">(</span><span class="n">root</span><span class="p">):</span>
<a id="__codelineno-7-2" name="__codelineno-7-2" href="#__codelineno-7-2"></a> <span class="k">if</span> <span class="ow">not</span> <span class="n">root</span><span class="p">:</span> <span class="c1"># 基本情况：空树高度为 0</span>
<a id="__codelineno-7-3" name="__codelineno-7-3" href="#__codelineno-7-3"></a> <span class="k">return</span> <span class="mi">0</span>
<a id="__codelineno-7-4" name="__codelineno-7-4" href="#__codelineno-7-4"></a> <span class="n">left_h</span> <span class="o">=</span> <span class="n">height</span><span class="p">(</span><span class="n">root</span><span class="o">.</span><span class="n">left</span><span class="p">)</span> <span class="c1"># 左子树高度</span>
<a id="__codelineno-7-5" name="__codelineno-7-5" href="#__codelineno-7-5"></a> <span class="n">right_h</span> <span class="o">=</span> <span class="n">height</span><span class="p">(</span><span class="n">root</span><span class="o">.</span><span class="n">right</span><span class="p">)</span> <span class="c1"># 右子树高度</span>
<a id="__codelineno-7-6" name="__codelineno-7-6" href="#__codelineno-7-6"></a> <span class="k">return</span> <span class="mi">1</span> <span class="o">+</span> <span class="nb">max</span><span class="p">(</span><span class="n">left_h</span><span class="p">,</span> <span class="n">right_h</span><span class="p">)</span> <span class="c1"># 当前节点增加 1 层</span>
</code></pre></div>
<ul>
<li>这种模式——"递归左子树，递归右子树，合并结果"——解决了绝大多数树的问题（见文件03）。</li>
</ul>
<h3 id="vs">递归 vs 迭代<a class="headerlink" href="#vs" title="Permanent link">&para;</a></h3>
<ul>
<li>
<p>每个递归算法都可以转换为迭代算法（使用显式栈或循环）。迭代避免了调用栈开销和栈溢出风险。</p>
</li>
<li>
<p><strong>何时优先使用递归</strong>：问题具有自然的递归结构（树、嵌套数据、分治法）。递归解法更简洁、更易于推理。</p>
</li>
<li>
<p><strong>何时优先使用迭代</strong>：递归深度可能非常大（例如，处理包含 <span class="arithmatex">\(10^6\)</span> 个节点的链表）。迭代解法避免了栈溢出。</p>
</li>
<li>
<p><strong>尾递归</strong>：如果递归调用是函数中的最后一个操作（递归调用返回后没有后续工作），则该递归调用是"尾递归"的。某些语言（Scheme、Scala）会将尾调用优化为使用常数栈空间。Python <strong>不</strong>优化尾调用，因此 Python 中的尾递归仍然使用 <span class="arithmatex">\(O(n)\)</span> 栈空间。</p>
</li>
</ul>
<h3 id="_10">常见陷阱<a class="headerlink" href="#_10" title="Permanent link">&para;</a></h3>
<table>
<thead>
<tr>
<th>陷阱</th>
<th>示例</th>
<th>修复</th>
</tr>
</thead>
<tbody>
<tr>
<td>缺少基本情况</td>
<td>无限递归 → 栈溢出</td>
<td>始终定义何时停止</td>
</tr>
<tr>
<td>基本情况错误</td>
<td>递归分解中的差一错误</td>
<td>用最小的输入测试（0、1、2）</td>
</tr>
<tr>
<td>问题规模未减小</td>
<td><code>f(n)</code> 调用 <code>f(n)</code> 而非 <code>f(n-1)</code></td>
<td>确保子问题严格更小</td>
</tr>
<tr>
<td>冗余计算</td>
<td>斐波那契数列：<code>f(n) = f(n-1) + f(n-2)</code> 以指数级重复计算</td>
<td>使用记忆化（→ DP）</td>
</tr>
<tr>
<td>Python 递归限制</td>
<td><code>factorial(10000)</code> 崩溃</td>
<td>使用 <code>sys.setrecursionlimit</code> 或转为迭代</td>
</tr>
</tbody>
</table>
<hr />
<h2 id="_11">回溯<a class="headerlink" href="#_11" title="Permanent link">&para;</a></h2>
<ul>
<li>
<p><strong>回溯</strong>是一种系统地探索所有可能解法的方法，通过逐步构建解并在发现部分解不可能得到有效答案时立即放弃。</p>
</li>
<li>
<p>可以把它想象成走迷宫。在每个岔路口，你选择一条路。如果碰到死胡同，你就回到上一个岔路口尝试不同的路。你不会从头开始——你<strong>回溯</strong>到最近的一个决策点。</p>
</li>
</ul>
<h3 id="_12">三个步骤<a class="headerlink" href="#_12" title="Permanent link">&para;</a></h3>
<p>每个回溯算法都遵循相同的模式：</p>
<ol>
<li><strong>选择</strong>：选择一个候选来扩展当前的部分解。</li>
<li><strong>探索</strong>：递归地尝试从这个候选构建一个完整的解。</li>
<li><strong>撤销</strong>：撤销选择（回溯）并尝试下一个候选。</li>
</ol>
<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">backtrack</span><span class="p">(</span><span class="n">state</span><span class="p">,</span> <span class="n">choices</span><span class="p">,</span> <span class="n">result</span><span class="p">):</span>
<a id="__codelineno-8-2" name="__codelineno-8-2" href="#__codelineno-8-2"></a> <span class="k">if</span> <span class="n">is_complete</span><span class="p">(</span><span class="n">state</span><span class="p">):</span>
<a id="__codelineno-8-3" name="__codelineno-8-3" href="#__codelineno-8-3"></a> <span class="n">result</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">state</span><span class="o">.</span><span class="n">copy</span><span class="p">())</span>
<a id="__codelineno-8-4" name="__codelineno-8-4" href="#__codelineno-8-4"></a> <span class="k">return</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">choice</span> <span class="ow">in</span> <span class="n">choices</span><span class="p">:</span>
<a id="__codelineno-8-7" name="__codelineno-8-7" href="#__codelineno-8-7"></a> <span class="k">if</span> <span class="n">is_valid</span><span class="p">(</span><span class="n">choice</span><span class="p">,</span> <span class="n">state</span><span class="p">):</span>
<a id="__codelineno-8-8" name="__codelineno-8-8" href="#__codelineno-8-8"></a> <span class="n">state</span><span class="o">.</span><span class="n">add</span><span class="p">(</span><span class="n">choice</span><span class="p">)</span> <span class="c1"># 1. 选择</span>
<a id="__codelineno-8-9" name="__codelineno-8-9" href="#__codelineno-8-9"></a> <span class="n">backtrack</span><span class="p">(</span><span class="n">state</span><span class="p">,</span> <span class="n">choices</span><span class="p">,</span> <span class="n">result</span><span class="p">)</span> <span class="c1"># 2. 探索</span>
<a id="__codelineno-8-10" name="__codelineno-8-10" href="#__codelineno-8-10"></a> <span class="n">state</span><span class="o">.</span><span class="n">remove</span><span class="p">(</span><span class="n">choice</span><span class="p">)</span> <span class="c1"># 3. 撤销（回溯）</span>
</code></pre></div>
<ul>
<li><strong>撤销</strong>步骤是回溯与普通递归的区别所在。没有它，状态会累积所有选择，你就无法探索替代路径。</li>
</ul>
<h3 id="_13">何时使用回溯<a class="headerlink" href="#_13" title="Permanent link">&para;</a></h3>
<ul>
<li>问题要求<strong>枚举所有有效配置</strong>：所有排列、所有子集、所有有效排列（如 N 皇后）。</li>
<li>问题要求<strong>寻找任何有效配置</strong>：数独求解、迷宫寻路。</li>
<li>搜索空间很大但可以<strong>剪枝</strong>：大多数部分解可以在完全探索之前被提前拒绝。</li>
</ul>
<h3 id="_14">剪枝如何使其变快<a class="headerlink" href="#_14" title="Permanent link">&para;</a></h3>
<ul>
<li>没有剪枝时，回溯会探索所有可能的组合——指数级时间。<strong>剪枝</strong>则提前砍掉分支：</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">for</span> <span class="n">choice</span> <span class="ow">in</span> <span class="n">choices</span><span class="p">:</span>
<a id="__codelineno-9-2" name="__codelineno-9-2" href="#__codelineno-9-2"></a> <span class="k">if</span> <span class="ow">not</span> <span class="n">is_valid</span><span class="p">(</span><span class="n">choice</span><span class="p">,</span> <span class="n">state</span><span class="p">):</span>
<a id="__codelineno-9-3" name="__codelineno-9-3" href="#__codelineno-9-3"></a> <span class="k">continue</span> <span class="c1"># 剪枝：跳过整个子树</span>
<a id="__codelineno-9-4" name="__codelineno-9-4" href="#__codelineno-9-4"></a>
<a id="__codelineno-9-5" name="__codelineno-9-5" href="#__codelineno-9-5"></a> <span class="n">state</span><span class="o">.</span><span class="n">add</span><span class="p">(</span><span class="n">choice</span><span class="p">)</span>
<a id="__codelineno-9-6" name="__codelineno-9-6" href="#__codelineno-9-6"></a> <span class="n">backtrack</span><span class="p">(</span><span class="n">state</span><span class="p">,</span> <span class="n">choices</span><span class="p">,</span> <span class="n">result</span><span class="p">)</span>
<a id="__codelineno-9-7" name="__codelineno-9-7" href="#__codelineno-9-7"></a> <span class="n">state</span><span class="o">.</span><span class="n">remove</span><span class="p">(</span><span class="n">choice</span><span class="p">)</span>
</code></pre></div>
<ul>
<li>在 N 皇后问题（文件05）中，在放置皇后之前检查列和对角线冲突，将搜索树从 <span class="arithmatex">\(n^n\)</span> 剪枝到大约 <span class="arithmatex">\(n!\)</span> 个候选。对于 <span class="arithmatex">\(n = 8\)</span>，这是 1600 万 → 40,000。好的剪枝使指数级算法在中等规模的 <span class="arithmatex">\(n\)</span> 下变得可行。</li>
</ul>
<h3 id="_15">生成所有子集（最简单的回溯）<a class="headerlink" href="#_15" title="Permanent link">&para;</a></h3>
<div class="highlight"><pre><span></span><code><a id="__codelineno-10-1" name="__codelineno-10-1" href="#__codelineno-10-1"></a><span class="k">def</span><span class="w"> </span><span class="nf">subsets</span><span class="p">(</span><span class="n">nums</span><span class="p">):</span>
<a id="__codelineno-10-2" name="__codelineno-10-2" href="#__codelineno-10-2"></a> <span class="n">result</span> <span class="o">=</span> <span class="p">[]</span>
<a id="__codelineno-10-3" name="__codelineno-10-3" href="#__codelineno-10-3"></a>
<a id="__codelineno-10-4" name="__codelineno-10-4" href="#__codelineno-10-4"></a> <span class="k">def</span><span class="w"> </span><span class="nf">backtrack</span><span class="p">(</span><span class="n">start</span><span class="p">,</span> <span class="n">path</span><span class="p">):</span>
<a id="__codelineno-10-5" name="__codelineno-10-5" href="#__codelineno-10-5"></a> <span class="n">result</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">path</span><span class="p">[:])</span> <span class="c1"># 每个部分解都是一个有效的子集</span>
<a id="__codelineno-10-6" name="__codelineno-10-6" href="#__codelineno-10-6"></a>
<a id="__codelineno-10-7" name="__codelineno-10-7" href="#__codelineno-10-7"></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">start</span><span class="p">,</span> <span class="nb">len</span><span class="p">(</span><span class="n">nums</span><span class="p">)):</span>
<a id="__codelineno-10-8" name="__codelineno-10-8" href="#__codelineno-10-8"></a> <span class="n">path</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">nums</span><span class="p">[</span><span class="n">i</span><span class="p">])</span> <span class="c1"># 选择</span>
<a id="__codelineno-10-9" name="__codelineno-10-9" href="#__codelineno-10-9"></a> <span class="n">backtrack</span><span class="p">(</span><span class="n">i</span> <span class="o">+</span> <span class="mi">1</span><span class="p">,</span> <span class="n">path</span><span class="p">)</span> <span class="c1"># 探索（i+1：不允许重复使用）</span>
<a id="__codelineno-10-10" name="__codelineno-10-10" href="#__codelineno-10-10"></a> <span class="n">path</span><span class="o">.</span><span class="n">pop</span><span class="p">()</span> <span class="c1"># 撤销</span>
<a id="__codelineno-10-11" name="__codelineno-10-11" href="#__codelineno-10-11"></a>
<a id="__codelineno-10-12" name="__codelineno-10-12" href="#__codelineno-10-12"></a> <span class="n">backtrack</span><span class="p">(</span><span class="mi">0</span><span class="p">,</span> <span class="p">[])</span>
<a id="__codelineno-10-13" name="__codelineno-10-13" href="#__codelineno-10-13"></a> <span class="k">return</span> <span class="n">result</span>
</code></pre></div>
<ul>
<li>
<p>对于 <code>[1, 2, 3]</code>，递归树：</p>
<ul>
<li><code>[]</code> → <code>[1]</code> → <code>[1,2]</code> → <code>[1,2,3]</code>（回溯）→ <code>[1,3]</code>（回溯）→ <code>[2]</code> → <code>[2,3]</code>（回溯）→ <code>[3]</code></li>
</ul>
</li>
<li>
<p>树中的每个节点是一次对 <code>backtrack</code> 的调用。每个叶子节点（以及中间节点）产生一个子集。总子集数：<span class="arithmatex">\(2^n\)</span>。</p>
</li>
</ul>
<h3 id="_16">生成所有排列<a class="headerlink" href="#_16" title="Permanent link">&para;</a></h3>
<div class="highlight"><pre><span></span><code><a id="__codelineno-11-1" name="__codelineno-11-1" href="#__codelineno-11-1"></a><span class="k">def</span><span class="w"> </span><span class="nf">permutations</span><span class="p">(</span><span class="n">nums</span><span class="p">):</span>
<a id="__codelineno-11-2" name="__codelineno-11-2" href="#__codelineno-11-2"></a> <span class="n">result</span> <span class="o">=</span> <span class="p">[]</span>
<a id="__codelineno-11-3" name="__codelineno-11-3" href="#__codelineno-11-3"></a>
<a id="__codelineno-11-4" name="__codelineno-11-4" href="#__codelineno-11-4"></a> <span class="k">def</span><span class="w"> </span><span class="nf">backtrack</span><span class="p">(</span><span class="n">path</span><span class="p">,</span> <span class="n">remaining</span><span class="p">):</span>
<a id="__codelineno-11-5" name="__codelineno-11-5" href="#__codelineno-11-5"></a> <span class="k">if</span> <span class="ow">not</span> <span class="n">remaining</span><span class="p">:</span>
<a id="__codelineno-11-6" name="__codelineno-11-6" href="#__codelineno-11-6"></a> <span class="n">result</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">path</span><span class="p">[:])</span>
<a id="__codelineno-11-7" name="__codelineno-11-7" href="#__codelineno-11-7"></a> <span class="k">return</span>
<a id="__codelineno-11-8" name="__codelineno-11-8" href="#__codelineno-11-8"></a>
<a id="__codelineno-11-9" name="__codelineno-11-9" href="#__codelineno-11-9"></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="nb">len</span><span class="p">(</span><span class="n">remaining</span><span class="p">)):</span>
<a id="__codelineno-11-10" name="__codelineno-11-10" href="#__codelineno-11-10"></a> <span class="n">path</span><span class="o">.</span><span class="n">append</span><span class="p">(</span><span class="n">remaining</span><span class="p">[</span><span class="n">i</span><span class="p">])</span> <span class="c1"># 选择</span>
<a id="__codelineno-11-11" name="__codelineno-11-11" href="#__codelineno-11-11"></a> <span class="n">backtrack</span><span class="p">(</span><span class="n">path</span><span class="p">,</span> <span class="n">remaining</span><span class="p">[:</span><span class="n">i</span><span class="p">]</span> <span class="o">+</span> <span class="n">remaining</span><span class="p">[</span><span class="n">i</span><span class="o">+</span><span class="mi">1</span><span class="p">:])</span> <span class="c1"># 探索</span>
<a id="__codelineno-11-12" name="__codelineno-11-12" href="#__codelineno-11-12"></a> <span class="n">path</span><span class="o">.</span><span class="n">pop</span><span class="p">()</span> <span class="c1"># 撤销</span>
<a id="__codelineno-11-13" name="__codelineno-11-13" href="#__codelineno-11-13"></a>
<a id="__codelineno-11-14" name="__codelineno-11-14" href="#__codelineno-11-14"></a> <span class="n">backtrack</span><span class="p">([],</span> <span class="n">nums</span><span class="p">)</span>
<a id="__codelineno-11-15" name="__codelineno-11-15" href="#__codelineno-11-15"></a> <span class="k">return</span> <span class="n">result</span>
</code></pre></div>
<ul>
<li>总排列数：<span class="arithmatex">\(n!\)</span>。每个排列需要 <span class="arithmatex">\(O(n)\)</span> 工作来构造 <code>remaining</code>，所以总复杂度为 <span class="arithmatex">\(O(n \cdot n!)\)</span>。</li>
</ul>
<h3 id="_17">常见陷阱<a class="headerlink" href="#_17" title="Permanent link">&para;</a></h3>
<table>
<thead>
<tr>
<th>陷阱</th>
<th>示例</th>
<th>修复</th>
</tr>
</thead>
<tbody>
<tr>
<td>忘记复制路径</td>
<td><code>result.append(path)</code> —— 所有条目共享同一个列表</td>
<td><code>result.append(path[:])</code> 或 <code>path.copy()</code></td>
</tr>
<tr>
<td>未回溯（撤销）</td>
<td>状态不断增长，后面的候选看到过时的状态</td>
<td>递归调用后始终执行 <code>path.pop()</code> 或 <code>state.remove()</code></td>
</tr>
<tr>
<td>循环起始位置错误</td>
<td>子集中有重复项，或排列中出现了不应有的重复使用</td>
<td>使用 <code>start</code> 参数避免重新访问之前的索引</td>
</tr>
<tr>
<td>跳过剪枝</td>
<td>探索明显无效的分支</td>
<td>在递归调用前添加 <code>if not is_valid: continue</code></td>
</tr>
</tbody>
</table>
<hr />
<h2 id="_18">动态规划<a class="headerlink" href="#_18" title="Permanent link">&para;</a></h2>
<ul>
<li>
<p><strong>动态规划（DP）</strong>是一种优化技术，适用于相同子问题被反复求解的情况。DP 不重复计算，而是每个子问题只解一次并存储结果。</p>
</li>
<li>
<p>DP 适用于具有两个性质的问题：</p>
<ol>
<li><strong>最优子结构</strong>：最优解可以由子问题的最优解构建而成。</li>
<li><strong>重叠子问题</strong>：相同的子问题在递归中多次出现。</li>
</ol>
</li>
</ul>
<h3 id="_19">斐波那契数列的动机<a class="headerlink" href="#_19" title="Permanent link">&para;</a></h3>
<ul>
<li>朴素递归斐波那契数列：</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="k">def</span><span class="w"> </span><span class="nf">fib</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
<a id="__codelineno-12-2" name="__codelineno-12-2" href="#__codelineno-12-2"></a> <span class="k">if</span> <span class="n">n</span> <span class="o">&lt;=</span> <span class="mi">1</span><span class="p">:</span>
<a id="__codelineno-12-3" name="__codelineno-12-3" href="#__codelineno-12-3"></a> <span class="k">return</span> <span class="n">n</span>
<a id="__codelineno-12-4" name="__codelineno-12-4" href="#__codelineno-12-4"></a> <span class="k">return</span> <span class="n">fib</span><span class="p">(</span><span class="n">n</span> <span class="o">-</span> <span class="mi">1</span><span class="p">)</span> <span class="o">+</span> <span class="n">fib</span><span class="p">(</span><span class="n">n</span> <span class="o">-</span> <span class="mi">2</span><span class="p">)</span>
</code></pre></div>
<ul>
<li>
<p>对于 <code>fib(5)</code>，递归树：</p>
<ul>
<li><code>fib(5)</code> 调用 <code>fib(4)</code> 和 <code>fib(3)</code></li>
<li><code>fib(4)</code> 调用 <code>fib(3)</code> 和 <code>fib(2)</code></li>
<li><code>fib(3)</code> 被计算了<strong>两次</strong>，<code>fib(2)</code> 被计算了<strong>三次</strong></li>
</ul>
</li>
<li>
<p>这是 <span class="arithmatex">\(O(2^n)\)</span>，因为树在每一层都分支，而且大多数分支重复计算相同的值。对于 <code>fib(50)</code>，需要超过 <span class="arithmatex">\(10^{15}\)</span> 次操作——不可行。</p>
</li>
<li>
<p>使用<strong>记忆化</strong>（自顶向下 DP）：</p>
</li>
</ul>
<div class="highlight"><pre><span></span><code><a id="__codelineno-13-1" name="__codelineno-13-1" href="#__codelineno-13-1"></a><span class="k">def</span><span class="w"> </span><span class="nf">fib_memo</span><span class="p">(</span><span class="n">n</span><span class="p">,</span> <span class="n">memo</span><span class="o">=</span><span class="p">{}):</span>
<a id="__codelineno-13-2" name="__codelineno-13-2" href="#__codelineno-13-2"></a> <span class="k">if</span> <span class="n">n</span> <span class="ow">in</span> <span class="n">memo</span><span class="p">:</span>
<a id="__codelineno-13-3" name="__codelineno-13-3" href="#__codelineno-13-3"></a> <span class="k">return</span> <span class="n">memo</span><span class="p">[</span><span class="n">n</span><span class="p">]</span>
<a id="__codelineno-13-4" name="__codelineno-13-4" href="#__codelineno-13-4"></a> <span class="k">if</span> <span class="n">n</span> <span class="o">&lt;=</span> <span class="mi">1</span><span class="p">:</span>
<a id="__codelineno-13-5" name="__codelineno-13-5" href="#__codelineno-13-5"></a> <span class="k">return</span> <span class="n">n</span>
<a id="__codelineno-13-6" name="__codelineno-13-6" href="#__codelineno-13-6"></a> <span class="n">memo</span><span class="p">[</span><span class="n">n</span><span class="p">]</span> <span class="o">=</span> <span class="n">fib_memo</span><span class="p">(</span><span class="n">n</span> <span class="o">-</span> <span class="mi">1</span><span class="p">,</span> <span class="n">memo</span><span class="p">)</span> <span class="o">+</span> <span class="n">fib_memo</span><span class="p">(</span><span class="n">n</span> <span class="o">-</span> <span class="mi">2</span><span class="p">,</span> <span class="n">memo</span><span class="p">)</span>
<a id="__codelineno-13-7" name="__codelineno-13-7" href="#__codelineno-13-7"></a> <span class="k">return</span> <span class="n">memo</span><span class="p">[</span><span class="n">n</span><span class="p">]</span>
</code></pre></div>
<ul>
<li>
<p>现在 <code>fib(3)</code> 只计算一次，存储起来，后续调用直接查找。总计：<span class="arithmatex">\(O(n)\)</span> 时间，<span class="arithmatex">\(O(n)\)</span> 空间。</p>
</li>
<li>
<p>使用<strong>制表法</strong>（自底向上 DP）：</p>
</li>
</ul>
<div class="highlight"><pre><span></span><code><a id="__codelineno-14-1" name="__codelineno-14-1" href="#__codelineno-14-1"></a><span class="k">def</span><span class="w"> </span><span class="nf">fib_tab</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
<a id="__codelineno-14-2" name="__codelineno-14-2" href="#__codelineno-14-2"></a> <span class="k">if</span> <span class="n">n</span> <span class="o">&lt;=</span> <span class="mi">1</span><span class="p">:</span>
<a id="__codelineno-14-3" name="__codelineno-14-3" href="#__codelineno-14-3"></a> <span class="k">return</span> <span class="n">n</span>
<a id="__codelineno-14-4" name="__codelineno-14-4" href="#__codelineno-14-4"></a> <span class="n">dp</span> <span class="o">=</span> <span class="p">[</span><span class="mi">0</span><span class="p">]</span> <span class="o">*</span> <span class="p">(</span><span class="n">n</span> <span class="o">+</span> <span class="mi">1</span><span class="p">)</span>
<a id="__codelineno-14-5" name="__codelineno-14-5" href="#__codelineno-14-5"></a> <span class="n">dp</span><span class="p">[</span><span class="mi">1</span><span class="p">]</span> <span class="o">=</span> <span class="mi">1</span>
<a id="__codelineno-14-6" name="__codelineno-14-6" href="#__codelineno-14-6"></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="mi">2</span><span class="p">,</span> <span class="n">n</span> <span class="o">+</span> <span class="mi">1</span><span class="p">):</span>
<a id="__codelineno-14-7" name="__codelineno-14-7" href="#__codelineno-14-7"></a> <span class="n">dp</span><span class="p">[</span><span class="n">i</span><span class="p">]</span> <span class="o">=</span> <span class="n">dp</span><span class="p">[</span><span class="n">i</span> <span class="o">-</span> <span class="mi">1</span><span class="p">]</span> <span class="o">+</span> <span class="n">dp</span><span class="p">[</span><span class="n">i</span> <span class="o">-</span> <span class="mi">2</span><span class="p">]</span>
<a id="__codelineno-14-8" name="__codelineno-14-8" href="#__codelineno-14-8"></a> <span class="k">return</span> <span class="n">dp</span><span class="p">[</span><span class="n">n</span><span class="p">]</span>
</code></pre></div>
<ul>
<li>同样 <span class="arithmatex">\(O(n)\)</span> 时间，但自底向上构建解，无需递归。可以进一步优化到 <span class="arithmatex">\(O(1)\)</span> 空间，因为每个值只依赖于前两个值。</li>
</ul>
<h3 id="dp">DP 配方<a class="headerlink" href="#dp" title="Permanent link">&para;</a></h3>
<p>对于任何 DP 问题，遵循以下步骤：</p>
<ol>
<li>
<p><strong>定义状态</strong>：<code>dp[i]</code>（或 <code>dp[i][j]</code>）代表什么？这是最难的一步。状态必须捕获足够的信息以做出最优决策。</p>
</li>
<li>
<p><strong>写出递推关系</strong>：<code>dp[i]</code> 如何与更小的子问题关联？这是转移公式。</p>
</li>
<li>
<p><strong>确定基本情况</strong>：哪些是最小的子问题，可以直接求解？</p>
</li>
<li>
<p><strong>确定迭代顺序</strong>：哪些子问题必须先于哪些子问题求解？自底向上：按照确保依赖关系已解决的顺序迭代。自顶向下：递归会自动处理。</p>
</li>
<li>
<p><strong>优化空间</strong>（可选）：如果 <code>dp[i]</code> 只依赖于前一行或前几个条目，你就不需要完整的表。</p>
</li>
</ol>
<h3 id="_20">示例：思路过程<a class="headerlink" href="#_20" title="Permanent link">&para;</a></h3>
<p><strong>问题</strong>：给定一个正整数数组，求不相邻元素的最大和（打家劫舍）。</p>
<p><strong>第1步——定义状态</strong>：<code>dp[i]</code> = 考虑元素 <code>nums[0..i]</code> 的最大和。</p>
<p><strong>第2步——写出递推关系</strong>：对于元素 <span class="arithmatex">\(i\)</span>，我们要么：
- 跳过它：<code>dp[i] = dp[i-1]</code>（不含元素 <span class="arithmatex">\(i\)</span> 的最佳和）。
- 取用它：<code>dp[i] = dp[i-2] + nums[i]</code>（必须跳过元素 <span class="arithmatex">\(i-1\)</span>，然后加上元素 <span class="arithmatex">\(i\)</span>）。</p>
<p>所以：<code>dp[i] = max(dp[i-1], dp[i-2] + nums[i])</code>。</p>
<p><strong>第3步——基本情况</strong>：<code>dp[0] = nums[0]</code>，<code>dp[1] = max(nums[0], nums[1])</code>。</p>
<p><strong>第4步——迭代顺序</strong>：从左到右（每个状态依赖于前两个状态）。</p>
<p><strong>第5步——空间优化</strong>：只需要最后两个值。</p>
<div class="highlight"><pre><span></span><code><a id="__codelineno-15-1" name="__codelineno-15-1" href="#__codelineno-15-1"></a><span class="k">def</span><span class="w"> </span><span class="nf">rob</span><span class="p">(</span><span class="n">nums</span><span class="p">):</span>
<a id="__codelineno-15-2" name="__codelineno-15-2" href="#__codelineno-15-2"></a> <span class="k">if</span> <span class="nb">len</span><span class="p">(</span><span class="n">nums</span><span class="p">)</span> <span class="o">==</span> <span class="mi">1</span><span class="p">:</span>
<a id="__codelineno-15-3" name="__codelineno-15-3" href="#__codelineno-15-3"></a> <span class="k">return</span> <span class="n">nums</span><span class="p">[</span><span class="mi">0</span><span class="p">]</span>
<a id="__codelineno-15-4" name="__codelineno-15-4" href="#__codelineno-15-4"></a>
<a id="__codelineno-15-5" name="__codelineno-15-5" href="#__codelineno-15-5"></a> <span class="n">prev2</span><span class="p">,</span> <span class="n">prev1</span> <span class="o">=</span> <span class="n">nums</span><span class="p">[</span><span class="mi">0</span><span class="p">],</span> <span class="nb">max</span><span class="p">(</span><span class="n">nums</span><span class="p">[</span><span class="mi">0</span><span class="p">],</span> <span class="n">nums</span><span class="p">[</span><span class="mi">1</span><span class="p">])</span>
<a id="__codelineno-15-6" name="__codelineno-15-6" href="#__codelineno-15-6"></a>
<a id="__codelineno-15-7" name="__codelineno-15-7" href="#__codelineno-15-7"></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="mi">2</span><span class="p">,</span> <span class="nb">len</span><span class="p">(</span><span class="n">nums</span><span class="p">)):</span>
<a id="__codelineno-15-8" name="__codelineno-15-8" href="#__codelineno-15-8"></a> <span class="n">curr</span> <span class="o">=</span> <span class="nb">max</span><span class="p">(</span><span class="n">prev1</span><span class="p">,</span> <span class="n">prev2</span> <span class="o">+</span> <span class="n">nums</span><span class="p">[</span><span class="n">i</span><span class="p">])</span>
<a id="__codelineno-15-9" name="__codelineno-15-9" href="#__codelineno-15-9"></a> <span class="n">prev2</span><span class="p">,</span> <span class="n">prev1</span> <span class="o">=</span> <span class="n">prev1</span><span class="p">,</span> <span class="n">curr</span>
<a id="__codelineno-15-10" name="__codelineno-15-10" href="#__codelineno-15-10"></a>
<a id="__codelineno-15-11" name="__codelineno-15-11" href="#__codelineno-15-11"></a> <span class="k">return</span> <span class="n">prev1</span>
</code></pre></div>
<h3 id="dp_1">如何识别 DP 问题<a class="headerlink" href="#dp_1" title="Permanent link">&para;</a></h3>
<ul>
<li>问题要求<strong>最优值</strong>（最小成本、最大利润、最长序列）或<strong>计数</strong>（方法数）。</li>
<li>问题在每一步都有<strong>选择</strong>（取/跳过、向左/向右、使用这枚硬币与否），并且整体最优答案依赖于子问题的最优答案。</li>
<li>画出递归树会显示<strong>重复的子问题</strong>。</li>
<li>暴力解法是指数级的，但<strong>不同的状态</strong>比递归调用少得多。</li>
</ul>
<h3 id="dp_2">DP 的分类<a class="headerlink" href="#dp_2" title="Permanent link">&para;</a></h3>
<ul>
<li>
<p><strong>1D DP</strong>：状态依赖于单个索引。示例：爬楼梯、打家劫舍、最大子数组。</p>
</li>
<li>
<p><strong>2D DP</strong>：状态依赖于两个索引。示例：最长公共子序列（<code>dp[i][j]</code> 表示字符串1的前 <span class="arithmatex">\(i\)</span> 个字符和字符串2的前 <span class="arithmatex">\(j\)</span> 个字符）、编辑距离、网格路径问题。</p>
</li>
<li>
<p><strong>区间 DP</strong>：状态是一个区间 <code>dp[i][j]</code>，表示 <code>arr[i..j]</code> 上的子问题。示例：矩阵链乘法、戳气球。</p>
</li>
<li>
<p><strong>背包 DP</strong>：状态是物品索引和容量。示例：0/1 背包、零钱兑换、子集和。</p>
</li>
<li>
<p><strong>位掩码 DP</strong>：状态包含一个位掩码，表示哪些元素已被使用。示例：旅行商问题、分配问题。状态空间为 <span class="arithmatex">\(O(2^n \cdot n)\)</span>，对于 <span class="arithmatex">\(n \leq 20\)</span> 可行。</p>
</li>
</ul>
<h3 id="vs_1">自顶向下 vs 自底向上<a class="headerlink" href="#vs_1" title="Permanent link">&para;</a></h3>
<table>
<thead>
<tr>
<th></th>
<th>自顶向下（记忆化）</th>
<th>自底向上（制表法）</th>
</tr>
</thead>
<tbody>
<tr>
<td>实现</td>
<td>递归 + 缓存</td>
<td>迭代 + 表</td>
</tr>
<tr>
<td>计算</td>
<td>只计算实际需要的子问题</td>
<td>计算直到目标的所有子问题</td>
</tr>
<tr>
<td>栈溢出风险</td>
<td>有（深度递归）</td>
<td>无</td>
</tr>
<tr>
<td>空间优化</td>
<td>较难</td>
<td>较易（使用滚动数组）</td>
</tr>
<tr>
<td>编码难度</td>
<td>通常更自然（写递归，加缓存）</td>
<td>需要考虑迭代顺序</td>
</tr>
</tbody>
</table>
<ul>
<li>在面试中，自顶向下通常编码更快。在生产环境中，自底向上通常更受青睐（无递归开销，缓存行为更好）。</li>
</ul>
<h3 id="_21">常见陷阱<a class="headerlink" href="#_21" title="Permanent link">&para;</a></h3>
<table>
<thead>
<tr>
<th>陷阱</th>
<th>示例</th>
<th>修复</th>
</tr>
</thead>
<tbody>
<tr>
<td>状态定义错误</td>
<td><code>dp[i]</code> 没有捕获足够信息来做决策</td>
<td>增加维度（例如用 <code>dp[i][j]</code> 代替 <code>dp[i]</code>）</td>
</tr>
<tr>
<td>缺少基本情况</td>
<td><code>dp[0]</code> 错误 → 所有后续值都错</td>
<td>手动验证基本情况</td>
</tr>
<tr>
<td>迭代顺序错误</td>
<td>在依赖关系未解决之前计算 <code>dp[i]</code></td>
<td>画出依赖箭头并相应迭代</td>
</tr>
<tr>
<td>未正确初始化 <code>dp</code></td>
<td>用 0 而应该用无穷大（求最小值时）</td>
<td>最小化用 <code>float('inf')</code>，最大化用 <code>float('-inf')</code></td>
</tr>
<tr>
<td>忘记考虑"跳过"选项</td>
<td>总是取当前元素</td>
<td>递推关系通常有 <code>max(take, skip)</code></td>
</tr>
<tr>
<td>可变的默认参数</td>
<td><code>def f(memo={})</code> 在调用间共享缓存</td>
<td><code>def f(memo=None): if memo is None: memo = {}</code></td>
</tr>
<tr>
<td>2D DP 中的差一错误</td>
<td><code>dp</code> 是 1-indexed 时访问 <code>text1[i]</code></td>
<td><code>dp</code> 大小为 <code>(m+1) x (n+1)</code>，访问 <code>text1[i-1]</code></td>
</tr>
</tbody>
</table>
<hr />
<h2 id="_22">融会贯通<a class="headerlink" href="#_22" title="Permanent link">&para;</a></h2>
<ul>
<li>
<p>这四个概念构成一个递进关系：</p>
<ol>
<li><strong>大O表示法</strong>告诉你一个方法是否足够快。</li>
<li><strong>递归</strong>将问题分解为子问题。</li>
<li><strong>回溯</strong>是递归 + 选择 + 撤销，用于穷举搜索。</li>
<li><strong>DP</strong>是递归 + 缓存，用于具有重叠子问题的优化。</li>
</ol>
</li>
<li>
<p>当你遇到一个新问题时：</p>
<ul>
<li>估计输入规模 <span class="arithmatex">\(n\)</span>。什么样的 Big O 是可接受的？</li>
<li>如果暴力解法是指数级的，且问题要求枚举/寻找配置：<strong>回溯</strong>（配合剪枝使其可行）。</li>
<li>如果暴力解法是指数级的，且问题要求最优值或计数，并且你看到重叠子问题：<strong>DP</strong>。</li>
<li>如果问题具有减半搜索空间的结构：<strong>二分查找</strong>或<strong>分治法</strong>。</li>
<li>如果问题涉及序列且有子数组约束：<strong>滑动窗口</strong>或<strong>双指针</strong>。</li>
<li>如果问题需要快速查找：<strong>哈希表</strong>。</li>
</ul>
</li>
</ul>
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