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<h1 id="121">12.1. &nbsp; 回溯算法<a class="headerlink" href="#121" title="Permanent link">&para;</a></h1>
<p>「回溯算法 Backtracking Algorithm」是一种通过穷举来解决问题的方法，它的核心思想是从一个初始状态出发，暴力搜索所有可能的解决方案，当遇到正确的解则将其记录，直到找到解或者尝试了所有可能的选择都无法找到解为止。</p>
<p>回溯算法通常采用「深度优先搜索」来遍历解空间。在二叉树章节中，我们提到前序、中序和后序遍历都属于深度优先搜索。下面，我们将从前序遍历入手，逐步了解回溯算法的工作原理。</p>
<p>回溯算法通常采用「深度优先搜索」来遍历解空间。在二叉树章节中，我们提到前序、中序和后序遍历都属于深度优先搜索。接下来我们先用前序遍历构造一个回溯问题，逐步了解回溯算法的工作原理。</p>
<div class="admonition question">
<p class="admonition-title">例题一</p>
<p>给定一个二叉树，搜索并记录所有值为 <span class="arithmatex">\(7\)</span> 的节点，返回节点列表。</p>
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<p>相较于基于前序遍历的实现代码，基于回溯算法框架的实现代码虽然显得啰嗦，但通用性更好。实际上，<strong>所有回溯问题都可以在该框架下解决</strong>。我们需要根据具体问题来定义 <code>state</code> 和 <code>choices</code> ，并实现框架中的各个方法。</p>
<h2 id="1215">12.1.5. &nbsp; 典型例题<a class="headerlink" href="#1215" title="Permanent link">&para;</a></h2>
<h2 id="1215">12.1.5. &nbsp; 优势与局限性<a class="headerlink" href="#1215" title="Permanent link">&para;</a></h2>
<p>回溯算法本质上是一种深度优先搜索算法，它尝试所有可能的解决方案直到找到满足条件的解。这种方法的优势在于它能够找到所有可能的解决方案，而且在合理的剪枝操作下，具有很高的效率。</p>
<p>然而，在处理大规模或者复杂问题时，<strong>回溯算法的运行效率可能难以接受</strong>。</p>
<ul>
<li>在最坏的情况下，回溯算法需要遍历解空间的所有可能解，所需时间很长。例如，求解 <span class="arithmatex">\(n\)</span> 皇后问题的时间复杂度可以达到 <span class="arithmatex">\(O(n!)\)</span> 。</li>
<li>在每一次递归调用时，都需要保存当前的状态（例如选择路径、用于剪枝的辅助变量等），对于深度很大的递归，空间需求可能会变得非常大。</li>
</ul>
<p>即便如此，<strong>回溯算法仍然是某些搜索问题和约束满足问题的最佳解决方案</strong>。对于这些问题，由于无法预测哪些选择可生成有效的解，因此我们必须对所有可能的选择进行遍历。在这种情况下，<strong>关键是如何进行效率优化</strong>：</p>
<ul>
<li>上文介绍过的剪枝是一种常用的优化方法。它可以避免搜索那些肯定不会产生有效解的路径，从而节省时间和空间。</li>
<li>另一个常用的优化方法是加入「启发式搜索 Heuristic Search」策略，它在搜索过程中引入一些策略或者估计值，从而优先搜索最有可能产生有效解的路径。</li>
</ul>
<h2 id="1216">12.1.6. &nbsp; 典型例题<a class="headerlink" href="#1216" title="Permanent link">&para;</a></h2>
<p><strong>搜索问题</strong>：这类问题的目标是找到满足特定条件的解决方案。</p>
<ul>
<li>全排列问题：给定一个集合，求出其所有可能的排列组合。</li>
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<li>最大团问题：给定一个无向图，找到最大的完全子图，即子图中的任意两个顶点之间都有边相连。</li>
</ul>
<p>请注意，回溯算法通常不是解决组合优化问题的最优方法。0-1 背包问题通常使用动态规划解决；旅行商是一个 NP-Hard 问题，常用解决方法有遗传算法和蚁群算法等；最大团问题是图轮中的一个经典 NP-Hard 问题，通常用贪心算法等启发式算法来解决。</p>
<h2 id="1216">12.1.6. &nbsp; 优势与局限性<a class="headerlink" href="#1216" title="Permanent link">&para;</a></h2>
<p>回溯算法本质上是一种深度优先搜索算法，它尝试所有可能的解决方案直到找到满足条件的解。这种方法的优势在于它能够找到所有可能的解决方案，而且在合理的剪枝操作下，具有很高的效率。</p>
<p>然而，在处理大规模或者复杂问题时，<strong>回溯算法的运行效率可能难以接受</strong>。这是因为在最坏的情况下，回溯算法需要遍历解空间的所有可能解。例如，求解 <span class="arithmatex">\(n\)</span> 皇后问题的时间复杂度可以达到 <span class="arithmatex">\(O(n!)\)</span> 。回溯算法的空间复杂度也可能较高。因为在每一次递归调用时，都需要保存当前的状态（例如选择路径、用于剪枝的辅助变量等），对于深度很大的递归，空间需求可能会变得非常大。</p>
<p>即便如此，<strong>回溯算法仍然是某些搜索问题和约束满足问题的最佳解决方案</strong>。对于这些问题，由于我们无法预测哪些选择可生成有效的解，因此我们必须对所有可能的选择进行遍历。在这种情况下，<strong>关键是如何进行效率优化</strong>：</p>
<ul>
<li>上文介绍过的剪枝是一种常用的优化方法。它可以避免搜索那些肯定不会产生有效解的路径，从而节省时间和空间。</li>
<li>另一个常用的优化方法是加入「启发式搜索 Heuristic Search」策略，它在搜索过程中引入一些策略或者估计值，从而优先搜索最有可能产生有效解的路径。</li>
</ul>

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<h1 id="122">12.2. &nbsp; 全排列问题<a class="headerlink" href="#122" title="Permanent link">&para;</a></h1>
<p>全排列问题是回溯算法的一个典型应用。它的定义是在给定一个集合（如一个数组或字符串）的情况下，找出这个集合中元素的所有可能的排列。</p>
<p>如下表所示，列举了几个示例数组和其对应的所有排列。</p>
<p>下表列举了几个示例数据，包括输入数组和对应的所有排列。</p>
<div class="center-table">
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<p>需要重点关注的是，我们引入了一个布尔型数组 <code>selected</code> ，它的长度与输入数组长度相等，其中 <code>selected[i]</code> 表示 <code>choices[i]</code> 是否已被选择。我们利用 <code>selected</code> 避免某个元素被重复选择，从而实现剪枝。</p>
<p>如下图所示，假设我们第一轮选择 1 ，第二轮选择 3 ，第三轮选择 2 ，则需要在第二轮剪掉元素 1 的分支，在第三轮剪掉元素 1, 3 的分支。<strong>从本质上理解，此剪枝操作可将搜索空间大小从 <span class="arithmatex">\(O(n^n)\)</span> 降低至 <span class="arithmatex">\(O(n!)\)</span></strong> 。</p>
<p>如下图所示，假设我们第一轮选择 1 ，第二轮选择 3 ，第三轮选择 2 ，则需要在第二轮剪掉元素 1 的分支，在第三轮剪掉元素 1, 3 的分支。<strong>此剪枝操作可将搜索空间大小从 <span class="arithmatex">\(O(n^n)\)</span> 降低至 <span class="arithmatex">\(O(n!)\)</span></strong> 。</p>
<p><img alt="全排列剪枝示例" src="../permutations_problem.assets/permutations_i_pruning.png" /></p>
<p align="center"> Fig. 全排列剪枝示例 </p>
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<p><img alt="重复排列" src="../permutations_problem.assets/permutations_ii.png" /></p>
<p align="center"> Fig. 重复排列 </p>
<p>那么，如何去除重复的排列呢？最直接地，我们可以借助一个哈希表，直接对排列结果进行去重。然而，这样做不够优雅，因为生成重复排列的搜索分支是没有必要的，应当被提前识别并剪枝，这样可以提升算法效率。</p>
<p>那么，如何去除重复的排列呢？最直接地，我们可以借助一个哈希表，直接对排列结果进行去重。然而这样做不够优雅，<strong>因为生成重复排列的搜索分支是没有必要的，应当被提前识别并剪枝</strong>，这样可以进一步提升算法效率。</p>
<p>观察发现，在第一轮中，选择 <span class="arithmatex">\(1\)</span> 或选择 <span class="arithmatex">\(\hat{1}\)</span> 是等价的，因为在这两个选择之下生成的所有排列都是重复的。因此，我们应该把 <span class="arithmatex">\(\hat{1}\)</span> 剪枝掉。同理，在第一轮选择 <span class="arithmatex">\(2\)</span> 后，第二轮选择中的 <span class="arithmatex">\(1\)</span> 和 <span class="arithmatex">\(\hat{1}\)</span> 也会产生重复分支，因此也需要将第二轮的 <span class="arithmatex">\(\hat{1}\)</span> 剪枝。</p>
<p><img alt="重复排列剪枝" src="../permutations_problem.assets/permutations_ii_pruning.png" /></p>
<p align="center"> Fig. 重复排列剪枝 </p>

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<div class="tabbed-set tabbed-alternate" data-tabs="1:11"><input checked="checked" id="__tabbed_1_1" name="__tabbed_1" type="radio" /><input id="__tabbed_1_2" name="__tabbed_1" type="radio" /><input id="__tabbed_1_3" name="__tabbed_1" type="radio" /><input id="__tabbed_1_4" name="__tabbed_1" type="radio" /><input id="__tabbed_1_5" name="__tabbed_1" type="radio" /><input id="__tabbed_1_6" name="__tabbed_1" type="radio" /><input id="__tabbed_1_7" name="__tabbed_1" type="radio" /><input id="__tabbed_1_8" name="__tabbed_1" type="radio" /><input id="__tabbed_1_9" name="__tabbed_1" type="radio" /><input id="__tabbed_1_10" name="__tabbed_1" type="radio" /><input id="__tabbed_1_11" name="__tabbed_1" type="radio" /><div class="tabbed-labels"><label for="__tabbed_1_1">Java</label><label for="__tabbed_1_2">C++</label><label for="__tabbed_1_3">Python</label><label for="__tabbed_1_4">Go</label><label for="__tabbed_1_5">JavaScript</label><label for="__tabbed_1_6">TypeScript</label><label for="__tabbed_1_7">C</label><label for="__tabbed_1_8">C#</label><label for="__tabbed_1_9">Swift</label><label for="__tabbed_1_10">Zig</label><label for="__tabbed_1_11">Dart</label></div>
<div class="tabbed-content">
<div class="tabbed-block">
<div class="highlight"><pre><span></span><code><a id="__codelineno-0-1" name="__codelineno-0-1" href="#__codelineno-0-1"></a><span class="cm">/* 使用多种「基本数据类型」来初始化「数组」 */</span>
<div class="highlight"><pre><span></span><code><a id="__codelineno-0-1" name="__codelineno-0-1" href="#__codelineno-0-1"></a><span class="c1">// 使用多种基本数据类型来初始化数组</span>
<a id="__codelineno-0-2" name="__codelineno-0-2" href="#__codelineno-0-2"></a><span class="kt">int</span><span class="o">[]</span><span class="w"> </span><span class="n">numbers</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="k">new</span><span class="w"> </span><span class="kt">int</span><span class="o">[</span><span class="mi">5</span><span class="o">]</span><span class="p">;</span>
<a id="__codelineno-0-3" name="__codelineno-0-3" href="#__codelineno-0-3"></a><span class="kt">float</span><span class="o">[]</span><span class="w"> </span><span class="n">decimals</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="k">new</span><span class="w"> </span><span class="kt">float</span><span class="o">[</span><span class="mi">5</span><span class="o">]</span><span class="p">;</span>
<a id="__codelineno-0-4" name="__codelineno-0-4" href="#__codelineno-0-4"></a><span class="kt">char</span><span class="o">[]</span><span class="w"> </span><span class="n">characters</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="k">new</span><span class="w"> </span><span class="kt">char</span><span class="o">[</span><span class="mi">5</span><span class="o">]</span><span class="p">;</span>
@ -2268,7 +2284,7 @@
</code></pre></div>
</div>
<div class="tabbed-block">
<div class="highlight"><pre><span></span><code><a id="__codelineno-1-1" name="__codelineno-1-1" href="#__codelineno-1-1"></a><span class="cm">/* 使用多种「基本数据类型」来初始化「数组」 */</span>
<div class="highlight"><pre><span></span><code><a id="__codelineno-1-1" name="__codelineno-1-1" href="#__codelineno-1-1"></a><span class="c1">// 使用多种基本数据类型来初始化数组</span>
<a id="__codelineno-1-2" name="__codelineno-1-2" href="#__codelineno-1-2"></a><span class="kt">int</span><span class="w"> </span><span class="n">numbers</span><span class="p">[</span><span class="mi">5</span><span class="p">];</span>
<a id="__codelineno-1-3" name="__codelineno-1-3" href="#__codelineno-1-3"></a><span class="kt">float</span><span class="w"> </span><span class="n">decimals</span><span class="p">[</span><span class="mi">5</span><span class="p">];</span>
<a id="__codelineno-1-4" name="__codelineno-1-4" href="#__codelineno-1-4"></a><span class="kt">char</span><span class="w"> </span><span class="n">characters</span><span class="p">[</span><span class="mi">5</span><span class="p">];</span>
@ -2289,19 +2305,19 @@
</code></pre></div>
</div>
<div class="tabbed-block">
<div class="highlight"><pre><span></span><code><a id="__codelineno-4-1" name="__codelineno-4-1" href="#__codelineno-4-1"></a><span class="cm">/* JavaScript 的数组可以自由存储各种基本数据类型和对象 */</span>
<div class="highlight"><pre><span></span><code><a id="__codelineno-4-1" name="__codelineno-4-1" href="#__codelineno-4-1"></a><span class="c1">// JavaScript 的数组可以自由存储各种基本数据类型和对象</span>
<a id="__codelineno-4-2" name="__codelineno-4-2" href="#__codelineno-4-2"></a><span class="kd">const</span><span class="w"> </span><span class="nx">array</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="p">[</span><span class="mf">0</span><span class="p">,</span><span class="w"> </span><span class="mf">0.0</span><span class="p">,</span><span class="w"> </span><span class="s1">&#39;a&#39;</span><span class="p">,</span><span class="w"> </span><span class="kc">false</span><span class="p">];</span>
</code></pre></div>
</div>
<div class="tabbed-block">
<div class="highlight"><pre><span></span><code><a id="__codelineno-5-1" name="__codelineno-5-1" href="#__codelineno-5-1"></a><span class="cm">/* 使用多种「基本数据类型」来初始化「数组」 */</span>
<div class="highlight"><pre><span></span><code><a id="__codelineno-5-1" name="__codelineno-5-1" href="#__codelineno-5-1"></a><span class="c1">// 使用多种基本数据类型来初始化数组</span>
<a id="__codelineno-5-2" name="__codelineno-5-2" href="#__codelineno-5-2"></a><span class="kd">const</span><span class="w"> </span><span class="nx">numbers</span><span class="o">:</span><span class="w"> </span><span class="kt">number</span><span class="p">[]</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="p">[];</span>
<a id="__codelineno-5-3" name="__codelineno-5-3" href="#__codelineno-5-3"></a><span class="kd">const</span><span class="w"> </span><span class="nx">characters</span><span class="o">:</span><span class="w"> </span><span class="kt">string</span><span class="p">[]</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="p">[];</span>
<a id="__codelineno-5-4" name="__codelineno-5-4" href="#__codelineno-5-4"></a><span class="kd">const</span><span class="w"> </span><span class="nx">booleans</span><span class="o">:</span><span class="w"> </span><span class="kt">boolean</span><span class="p">[]</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="p">[];</span>
</code></pre></div>
</div>
<div class="tabbed-block">
<div class="highlight"><pre><span></span><code><a id="__codelineno-6-1" name="__codelineno-6-1" href="#__codelineno-6-1"></a><span class="cm">/* 使用多种「基本数据类型」来初始化「数组」 */</span>
<div class="highlight"><pre><span></span><code><a id="__codelineno-6-1" name="__codelineno-6-1" href="#__codelineno-6-1"></a><span class="c1">// 使用多种基本数据类型来初始化数组</span>
<a id="__codelineno-6-2" name="__codelineno-6-2" href="#__codelineno-6-2"></a><span class="kt">int</span><span class="w"> </span><span class="n">numbers</span><span class="p">[</span><span class="mi">10</span><span class="p">];</span>
<a id="__codelineno-6-3" name="__codelineno-6-3" href="#__codelineno-6-3"></a><span class="kt">float</span><span class="w"> </span><span class="n">decimals</span><span class="p">[</span><span class="mi">10</span><span class="p">];</span>
<a id="__codelineno-6-4" name="__codelineno-6-4" href="#__codelineno-6-4"></a><span class="kt">char</span><span class="w"> </span><span class="n">characters</span><span class="p">[</span><span class="mi">10</span><span class="p">];</span>
@ -2309,7 +2325,7 @@
</code></pre></div>
</div>
<div class="tabbed-block">
<div class="highlight"><pre><span></span><code><a id="__codelineno-7-1" name="__codelineno-7-1" href="#__codelineno-7-1"></a><span class="cm">/* 使用多种「基本数据类型」来初始化「数组」 */</span>
<div class="highlight"><pre><span></span><code><a id="__codelineno-7-1" name="__codelineno-7-1" href="#__codelineno-7-1"></a><span class="c1">// 使用多种基本数据类型来初始化数组</span>
<a id="__codelineno-7-2" name="__codelineno-7-2" href="#__codelineno-7-2"></a><span class="kt">int</span><span class="p">[]</span><span class="w"> </span><span class="n">numbers</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="k">new</span><span class="w"> </span><span class="kt">int</span><span class="p">[</span><span class="m">5</span><span class="p">];</span>
<a id="__codelineno-7-3" name="__codelineno-7-3" href="#__codelineno-7-3"></a><span class="kt">float</span><span class="p">[]</span><span class="w"> </span><span class="n">decimals</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="k">new</span><span class="w"> </span><span class="kt">float</span><span class="p">[</span><span class="m">5</span><span class="p">];</span>
<a id="__codelineno-7-4" name="__codelineno-7-4" href="#__codelineno-7-4"></a><span class="kt">char</span><span class="p">[]</span><span class="w"> </span><span class="n">characters</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="k">new</span><span class="w"> </span><span class="kt">char</span><span class="p">[</span><span class="m">5</span><span class="p">];</span>
@ -2317,7 +2333,7 @@
</code></pre></div>
</div>
<div class="tabbed-block">
<div class="highlight"><pre><span></span><code><a id="__codelineno-8-1" name="__codelineno-8-1" href="#__codelineno-8-1"></a><span class="cm">/* 使用多种「基本数据类型」来初始化「数组」 */</span>
<div class="highlight"><pre><span></span><code><a id="__codelineno-8-1" name="__codelineno-8-1" href="#__codelineno-8-1"></a><span class="c1">// 使用多种基本数据类型来初始化数组</span>
<a id="__codelineno-8-2" name="__codelineno-8-2" href="#__codelineno-8-2"></a><span class="kd">let</span> <span class="nv">numbers</span> <span class="p">=</span> <span class="nb">Array</span><span class="p">(</span><span class="n">repeating</span><span class="p">:</span> <span class="nb">Int</span><span class="p">(),</span> <span class="bp">count</span><span class="p">:</span> <span class="mi">5</span><span class="p">)</span>
<a id="__codelineno-8-3" name="__codelineno-8-3" href="#__codelineno-8-3"></a><span class="kd">let</span> <span class="nv">decimals</span> <span class="p">=</span> <span class="nb">Array</span><span class="p">(</span><span class="n">repeating</span><span class="p">:</span> <span class="nb">Double</span><span class="p">(),</span> <span class="bp">count</span><span class="p">:</span> <span class="mi">5</span><span class="p">)</span>
<a id="__codelineno-8-4" name="__codelineno-8-4" href="#__codelineno-8-4"></a><span class="kd">let</span> <span class="nv">characters</span> <span class="p">=</span> <span class="nb">Array</span><span class="p">(</span><span class="n">repeating</span><span class="p">:</span> <span class="nb">Character</span><span class="p">(</span><span class="s">&quot;a&quot;</span><span class="p">),</span> <span class="bp">count</span><span class="p">:</span> <span class="mi">5</span><span class="p">)</span>
@ -2329,7 +2345,7 @@
</code></pre></div>
</div>
<div class="tabbed-block">
<div class="highlight"><pre><span></span><code><a id="__codelineno-10-1" name="__codelineno-10-1" href="#__codelineno-10-1"></a><span class="cm">/* 使用多种「基本数据类型」来初始化「数组」 */</span>
<div class="highlight"><pre><span></span><code><a id="__codelineno-10-1" name="__codelineno-10-1" href="#__codelineno-10-1"></a><span class="c1">// 使用多种基本数据类型来初始化数组</span>
<a id="__codelineno-10-2" name="__codelineno-10-2" href="#__codelineno-10-2"></a><span class="n">List</span><span class="o">&lt;</span><span class="kt">int</span><span class="o">&gt;</span><span class="w"> </span><span class="n">numbers</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">List</span><span class="p">.</span><span class="n">filled</span><span class="p">(</span><span class="m">5</span><span class="p">,</span><span class="w"> </span><span class="m">0</span><span class="p">);</span>
<a id="__codelineno-10-3" name="__codelineno-10-3" href="#__codelineno-10-3"></a><span class="n">List</span><span class="o">&lt;</span><span class="kt">double</span><span class="o">&gt;</span><span class="w"> </span><span class="n">decimals</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">List</span><span class="p">.</span><span class="n">filled</span><span class="p">(</span><span class="m">5</span><span class="p">,</span><span class="w"> </span><span class="m">0.0</span><span class="p">);</span>
<a id="__codelineno-10-4" name="__codelineno-10-4" href="#__codelineno-10-4"></a><span class="n">List</span><span class="o">&lt;</span><span class="kt">String</span><span class="o">&gt;</span><span class="w"> </span><span class="n">characters</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">List</span><span class="p">.</span><span class="n">filled</span><span class="p">(</span><span class="m">5</span><span class="p">,</span><span class="w"> </span><span class="s1">&#39;a&#39;</span><span class="p">);</span>

16
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@ -1975,6 +1975,8 @@
@ -2053,6 +2055,20 @@
<li class="md-nav__item">
<a href="../../chapter_dynamic_programming/unbounded_knapsack_problem/" class="md-nav__link">
13.5. &nbsp; 完全背包问题（New）
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@ -1954,6 +1954,8 @@
@ -2032,6 +2034,20 @@
<li class="md-nav__item">
<a href="../../chapter_dynamic_programming/unbounded_knapsack_problem/" class="md-nav__link">
13.5. &nbsp; 完全背包问题（New）
</a>
</li>
</ul>
</nav>
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@ -2219,16 +2235,22 @@
<p>数据结构可以从逻辑结构和物理结构两个维度进行分类。</p>
<h2 id="311">3.1.1. &nbsp; 逻辑结构：线性与非线性<a class="headerlink" href="#311" title="Permanent link">&para;</a></h2>
<p><strong>「逻辑结构」揭示了数据元素之间的逻辑关系</strong>。在数组和链表中，数据按照顺序依次排列，体现了数据之间的线性关系；而在树中，数据从顶部向下按层次排列，表现出祖先与后代之间的派生关系；图则由节点和边构成，反映了复杂的网络关系。</p>
<p>逻辑结构通常分为「线性」和「非线性」两类。线性结构比较直观，指数据在逻辑关系上呈线性排列；非线性结构则相反，呈非线性排列，例如网状或树状结构。</p>
<p>逻辑结构通常分为“线性”和“非线性”两类。线性结构比较直观，指数据在逻辑关系上呈线性排列；非线性结构则相反，呈非线性排列。</p>
<ul>
<li><strong>线性数据结构</strong>：数组、链表、栈、队列、哈希表；</li>
<li><strong>非线性数据结构</strong>：树、图、堆、哈希表；</li>
<li><strong>非线性数据结构</strong>：树、堆、图、哈希表；</li>
</ul>
<p><img alt="线性与非线性数据结构" src="../classification_of_data_structure.assets/classification_logic_structure.png" /></p>
<p align="center"> Fig. 线性与非线性数据结构 </p>
<p>非线性数据结构可以进一步被划分为树形结构和网状结构。</p>
<ul>
<li><strong>线性结构</strong>：数组、链表、队列、栈、哈希表，元素存在一对一的顺序关系；</li>
<li><strong>树形结构</strong>：树、堆、哈希表，元素存在一对多的关系；</li>
<li><strong>网状结构</strong>：图，元素存在多对多的关系；</li>
</ul>
<h2 id="312">3.1.2. &nbsp; 物理结构：连续与离散<a class="headerlink" href="#312" title="Permanent link">&para;</a></h2>
<p>在计算机中，内存和硬盘是两种主要的存储硬件设备。「硬盘」主要用于长期存储数据，容量较大（通常可达到 TB 级别）、速度较慢。「内存」用于运行程序时暂存数据，速度较快，但容量较小（通常为 GB 级别）。</p>
<p>在计算机中，内存和硬盘是两种主要的存储硬件设备。硬盘主要用于长期存储数据，容量较大（通常可达到 TB 级别）、速度较慢。内存用于运行程序时暂存数据，速度较快，但容量较小（通常为 GB 级别）。</p>
<p><strong>在算法运行过程中，相关数据都存储在内存中</strong>。下图展示了一个计算机内存条，其中每个黑色方块都包含一块内存空间。我们可以将内存想象成一个巨大的 Excel 表格，其中每个单元格都可以存储 1 byte 的数据，在算法运行时，所有数据都被存储在这些单元格中。</p>
<p><strong>系统通过「内存地址 Memory Location」来访问目标内存位置的数据</strong>。计算机根据特定规则为表格中的每个单元格分配编号，确保每个内存空间都有唯一的内存地址。有了这些地址，程序便可以访问内存中的数据。</p>
<p><img alt="内存条、内存空间、内存地址" src="../classification_of_data_structure.assets/computer_memory_location.png" /></p>
@ -2244,7 +2266,7 @@
<li><strong>基于数组可实现</strong>：栈、队列、哈希表、树、堆、图、矩阵、张量（维度 <span class="arithmatex">\(\geq 3\)</span> 的数组）等；</li>
<li><strong>基于链表可实现</strong>：栈、队列、哈希表、树、堆、图等；</li>
</ul>
<p>基于数组实现的数据结构也被称为「静态数据结构」，这意味着此类数据结构在初始化后长度不可变。相对应地，基于链表实现的数据结构被称为「动态数据结构」，这类数据结构在初始化后，仍可以在程序运行过程中对其长度进行调整。</p>
<p>基于数组实现的数据结构也被称为“静态数据结构”，这意味着此类数据结构在初始化后长度不可变。相对应地，基于链表实现的数据结构被称为“动态数据结构”，这类数据结构在初始化后，仍可以在程序运行过程中对其长度进行调整。</p>
<div class="admonition tip">
<p class="admonition-title">Tip</p>
<p>如若感觉理解物理结构有困难，建议先阅读下一章“数组与链表”，然后再回头理解物理结构的含义。数组与链表是其他所有数据结构的基石，建议你投入更多时间深入了解这两种基本数据结构。</p>

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<a href="../../chapter_dynamic_programming/unbounded_knapsack_problem/" class="md-nav__link">
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<li class="md-nav__item">
<a href="../unbounded_knapsack_problem/" class="md-nav__link">
13.5. &nbsp; 完全背包问题（New）
</a>
</li>
</ul>
</nav>
</li>
@ -2225,7 +2241,7 @@
<p>给定一个楼梯，你每步可以上 <span class="arithmatex">\(1\)</span> 阶或者 <span class="arithmatex">\(2\)</span> 阶，每一阶楼梯上都贴有一个非负整数，表示你在该台阶所需要付出的代价。给定一个非负整数数组 <span class="arithmatex">\(cost\)</span> ，其中 <span class="arithmatex">\(cost[i]\)</span> 表示在第 <span class="arithmatex">\(i\)</span> 个台阶需要付出的代价，<span class="arithmatex">\(cost[0]\)</span> 为地面起始点。请计算最少需要付出多少代价才能到达顶部？</p>
</div>
<p>如下图所示，若第 <span class="arithmatex">\(1\)</span> , <span class="arithmatex">\(2\)</span> , <span class="arithmatex">\(3\)</span> 阶的代价分别为 <span class="arithmatex">\(1\)</span> , <span class="arithmatex">\(10\)</span> , <span class="arithmatex">\(1\)</span> ，则从地面爬到第 <span class="arithmatex">\(3\)</span> 阶的最小代价为 <span class="arithmatex">\(2\)</span> 。</p>
<p><img alt="爬到第 3 阶的最小代价" src="../intro_to_dynamic_programming.assets/min_cost_cs_example.png" /></p>
<p><img alt="爬到第 3 阶的最小代价" src="../dp_problem_features.assets/min_cost_cs_example.png" /></p>
<p align="center"> Fig. 爬到第 3 阶的最小代价 </p>
<p>设 <span class="arithmatex">\(dp[i]\)</span> 为爬到第 <span class="arithmatex">\(i\)</span> 阶累计付出的代价，由于第 <span class="arithmatex">\(i\)</span> 阶只可能从 <span class="arithmatex">\(i - 1\)</span> 阶或 <span class="arithmatex">\(i - 2\)</span> 阶走来，因此 <span class="arithmatex">\(dp[i]\)</span> 只可能等于 <span class="arithmatex">\(dp[i - 1] + cost[i]\)</span> 或 <span class="arithmatex">\(dp[i - 2] + cost[i]\)</span> 。为了尽可能减少代价，我们应该选择两者中较小的那一个，即：</p>
@ -2341,7 +2357,7 @@ dp[i] = \min(dp[i-1], dp[i-2]) + cost[i]
</div>
</div>
</div>
<p><img alt="爬楼梯最小代价的动态规划过程" src="../intro_to_dynamic_programming.assets/min_cost_cs_dp.png" /></p>
<p><img alt="爬楼梯最小代价的动态规划过程" src="../dp_problem_features.assets/min_cost_cs_dp.png" /></p>
<p align="center"> Fig. 爬楼梯最小代价的动态规划过程 </p>
<p>这道题同样也可以进行状态压缩，将一维压缩至零维，使得空间复杂度从 <span class="arithmatex">\(O(n)\)</span> 降低至 <span class="arithmatex">\(O(1)\)</span> 。</p>
@ -2446,7 +2462,7 @@ dp[i] = \min(dp[i-1], dp[i-2]) + cost[i]
<p>给定一个共有 <span class="arithmatex">\(n\)</span> 阶的楼梯，你每步可以上 <span class="arithmatex">\(1\)</span> 阶或者 <span class="arithmatex">\(2\)</span> 阶，<strong>但不能连续两轮跳 <span class="arithmatex">\(1\)</span> 阶</strong>，请问有多少种方案可以爬到楼顶。</p>
</div>
<p>例如，爬上第 <span class="arithmatex">\(3\)</span> 阶仅剩 <span class="arithmatex">\(2\)</span> 种可行方案，其中连续三次跳 <span class="arithmatex">\(1\)</span> 阶的方案不满足约束条件，因此被舍弃。</p>
<p><img alt="带约束爬到第 3 阶的方案数量" src="../intro_to_dynamic_programming.assets/climbing_stairs_constraint_example.png" /></p>
<p><img alt="带约束爬到第 3 阶的方案数量" src="../dp_problem_features.assets/climbing_stairs_constraint_example.png" /></p>
<p align="center"> Fig. 带约束爬到第 3 阶的方案数量 </p>
<p>在该问题中，<strong>下一步选择不能由当前状态（当前楼梯阶数）独立决定，还和前一个状态（上轮楼梯阶数）有关</strong>。如果上一轮是跳 <span class="arithmatex">\(1\)</span> 阶上来的，那么下一轮就必须跳 <span class="arithmatex">\(2\)</span> 阶。</p>
@ -2463,7 +2479,7 @@ dp[i, 1] = dp[i-1, 2] \\
dp[i, 2] = dp[i-2, 1] + dp[i-2, 2]
\end{cases}
\]</div>
<p><img alt="考虑约束下的递推关系" src="../intro_to_dynamic_programming.assets/climbing_stairs_constraint_state_transfer.png" /></p>
<p><img alt="考虑约束下的递推关系" src="../dp_problem_features.assets/climbing_stairs_constraint_state_transfer.png" /></p>
<p align="center"> Fig. 考虑约束下的递推关系 </p>
<p>最终，返回 <span class="arithmatex">\(dp[n, 1] + dp[n, 2]\)</span> 即可，两者之和代表爬到第 <span class="arithmatex">\(n\)</span> 阶的方案总数。</p>

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<li class="md-nav__item">
<a href="../unbounded_knapsack_problem/" class="md-nav__link">
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</a>
</li>
</ul>
</nav>
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@ -2325,12 +2341,12 @@ dp[i, j] = \min(dp[i-1, j], dp[i, j-1]) + grid[i, j]
<p class="admonition-title">Note</p>
<p>边界条件即初始状态，在搜索中用于剪枝，在动态规划中用于初始化 <span class="arithmatex">\(dp\)</span> 表。状态转移顺序的核心是要保证在计算当前问题时，所有它依赖的更小子问题都已经被正确地计算出来。</p>
</div>
<p>最后，我们基于以上结果实现解法即可。熟练度较高同学可以直接写出动态规划解法，初学者可以按照“暴力搜索 <span class="arithmatex">\(\rightarrow\)</span> 记忆化搜索 <span class="arithmatex">\(\rightarrow\)</span> 动态规划” 的顺序实现。</p>
<p>接下来，我们就可以实现动态规划代码了。然而，由于子问题分解是一种从顶至底的思想，因此按照“暴力搜索 <span class="arithmatex">\(\rightarrow\)</span> 记忆化搜索 <span class="arithmatex">\(\rightarrow\)</span> 动态规划”的顺序实现更加符合思维习惯。</p>
<h2 id="1333">13.3.3. &nbsp; 方法一：暴力搜索<a class="headerlink" href="#1333" title="Permanent link">&para;</a></h2>
<p>从状态 <span class="arithmatex">\([i, j]\)</span> 开始搜索，不断分解为更小的状态 <span class="arithmatex">\([i-1, j]\)</span> 和 <span class="arithmatex">\([i, j-1]\)</span> ，包括以下递归要素：</p>
<ul>
<li><strong>递归参数</strong>：状态 <span class="arithmatex">\([i, j]\)</span> ；<strong>返回值</strong>：从 <span class="arithmatex">\([0, 0]\)</span> 到 <span class="arithmatex">\([i, j]\)</span> 的最小路径和 <span class="arithmatex">\(dp[i, j]\)</span> ；</li>
<li><strong>终止条件</strong>：当 <span class="arithmatex">\(i = 0\)</span> 且 <span class="arithmatex">\(j = 0\)</span> 时，返回代价 <span class="arithmatex">\(grid[0][0]\)</span> ；</li>
<li><strong>终止条件</strong>：当 <span class="arithmatex">\(i = 0\)</span> 且 <span class="arithmatex">\(j = 0\)</span> 时，返回代价 <span class="arithmatex">\(grid[0, 0]\)</span> ；</li>
<li><strong>剪枝</strong>：当 <span class="arithmatex">\(i &lt; 0\)</span> 时或 <span class="arithmatex">\(j &lt; 0\)</span> 时索引越界，此时返回代价 <span class="arithmatex">\(+\infty\)</span> ，代表不可行；</li>
</ul>
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@ -18,7 +18,7 @@
<link rel="prev" href="../dp_solution_pipeline/">
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@ -2245,7 +2261,7 @@
<p><strong>第一步：思考每轮的决策，定义状态，从而得到 <span class="arithmatex">\(dp\)</span> 表</strong></p>
<p>在 0-1 背包问题中，不放入背包，背包容量不变；放入背包，背包容量减小。由此可得状态定义：当前物品编号 <span class="arithmatex">\(i\)</span> 和剩余背包容量 <span class="arithmatex">\(c\)</span> ，记为 <span class="arithmatex">\([i, c]\)</span> 。</p>
<p>状态 <span class="arithmatex">\([i, c]\)</span> 对应的子问题为：<strong>前 <span class="arithmatex">\(i\)</span> 个物品在剩余容量为 <span class="arithmatex">\(c\)</span> 的背包中的最大价值</strong>，记为 <span class="arithmatex">\(dp[i, c]\)</span> 。</p>
<p>至此，我们得到一个尺寸为 <span class="arithmatex">\(n \times cap\)</span> 的二维 <span class="arithmatex">\(dp\)</span> 矩阵。</p>
<p>需要求解的是 <span class="arithmatex">\(dp[n, cap]\)</span> ，因此需要一个尺寸为 <span class="arithmatex">\((n+1) \times (cap+1)\)</span> 的二维 <span class="arithmatex">\(dp\)</span> 表。</p>
<p><strong>第二步：找出最优子结构，进而推导出状态转移方程</strong></p>
<p>当我们做出物品 <span class="arithmatex">\(i\)</span> 的决策后，剩余的是前 <span class="arithmatex">\(i-1\)</span> 个物品的决策。因此，状态转移分为两种情况：</p>
<ul>
@ -2260,6 +2276,10 @@ dp[i, c] = \max(dp[i-1, c], dp[i-1, c - wgt[i-1]] + val[i-1])
<p><strong>第三步：确定边界条件和状态转移顺序</strong></p>
<p>当无物品或无剩余背包容量时最大价值为 <span class="arithmatex">\(0\)</span> ，即所有 <span class="arithmatex">\(dp[i, 0]\)</span> 和 <span class="arithmatex">\(dp[0, c]\)</span> 都等于 <span class="arithmatex">\(0\)</span> 。</p>
<p>当前状态 <span class="arithmatex">\([i, c]\)</span> 从上方的状态 <span class="arithmatex">\([i-1, c]\)</span> 和左上方的状态 <span class="arithmatex">\([i-1, c-wgt[i-1]]\)</span> 转移而来，因此通过两层循环正序遍历整个 <span class="arithmatex">\(dp\)</span> 表即可。</p>
<div class="admonition tip">
<p class="admonition-title">Tip</p>
<p>完成以上三步后，我们可以直接实现从底至顶的动态规划解法。而为了展示本题包含的重叠子问题，本文也同时给出从顶至底的暴力搜索和记忆化搜索解法。</p>
</div>
<h2 id="1341">13.4.1. &nbsp; 方法一：暴力搜索<a class="headerlink" href="#1341" title="Permanent link">&para;</a></h2>
<p>搜索代码包含以下要素：</p>
<ul>
@ -2510,7 +2530,7 @@ dp[i, c] = \max(dp[i-1, c], dp[i-1, c - wgt[i-1]] + val[i-1])
<p align="center"> Fig. 0-1 背包的记忆化搜索递归树 </p>
<h2 id="1343">13.4.3. &nbsp; 方法三：动态规划<a class="headerlink" href="#1343" title="Permanent link">&para;</a></h2>
<p>动态规划解法本质上就是在状态转移中填充 <code>dp</code> 矩阵的过程，代码如下所示。</p>
<p>动态规划解法本质上就是在状态转移中填充 <span class="arithmatex">\(dp\)</span> 表的过程，代码如下所示。</p>
<div class="tabbed-set tabbed-alternate" data-tabs="3:11"><input checked="checked" id="__tabbed_3_1" name="__tabbed_3" type="radio" /><input id="__tabbed_3_2" name="__tabbed_3" type="radio" /><input id="__tabbed_3_3" name="__tabbed_3" type="radio" /><input id="__tabbed_3_4" name="__tabbed_3" type="radio" /><input id="__tabbed_3_5" name="__tabbed_3" type="radio" /><input id="__tabbed_3_6" name="__tabbed_3" type="radio" /><input id="__tabbed_3_7" name="__tabbed_3" type="radio" /><input id="__tabbed_3_8" name="__tabbed_3" type="radio" /><input id="__tabbed_3_9" name="__tabbed_3" type="radio" /><input id="__tabbed_3_10" name="__tabbed_3" type="radio" /><input id="__tabbed_3_11" name="__tabbed_3" type="radio" /><div class="tabbed-labels"><label for="__tabbed_3_1">Java</label><label for="__tabbed_3_2">C++</label><label for="__tabbed_3_3">Python</label><label for="__tabbed_3_4">Go</label><label for="__tabbed_3_5">JavaScript</label><label for="__tabbed_3_6">TypeScript</label><label for="__tabbed_3_7">C</label><label for="__tabbed_3_8">C#</label><label for="__tabbed_3_9">Swift</label><label for="__tabbed_3_10">Zig</label><label for="__tabbed_3_11">Dart</label></div>
<div class="tabbed-content">
<div class="tabbed-block">
@ -2627,7 +2647,7 @@ dp[i, c] = \max(dp[i-1, c], dp[i-1, c - wgt[i-1]] + val[i-1])
</div>
</div>
</div>
<p>如下图所示，时间复杂度由 <code>dp</code> 矩阵大小决定，为 <span class="arithmatex">\(O(n \times cap)\)</span> 。</p>
<p>如下图所示，时间复杂度由数组 <code>dp</code> 大小决定，为 <span class="arithmatex">\(O(n \times cap)\)</span> 。</p>
<div class="tabbed-set tabbed-alternate" data-tabs="4:14"><input checked="checked" id="__tabbed_4_1" name="__tabbed_4" type="radio" /><input id="__tabbed_4_2" name="__tabbed_4" type="radio" /><input id="__tabbed_4_3" name="__tabbed_4" type="radio" /><input id="__tabbed_4_4" name="__tabbed_4" type="radio" /><input id="__tabbed_4_5" name="__tabbed_4" type="radio" /><input id="__tabbed_4_6" name="__tabbed_4" type="radio" /><input id="__tabbed_4_7" name="__tabbed_4" type="radio" /><input id="__tabbed_4_8" name="__tabbed_4" type="radio" /><input id="__tabbed_4_9" name="__tabbed_4" type="radio" /><input id="__tabbed_4_10" name="__tabbed_4" type="radio" /><input id="__tabbed_4_11" name="__tabbed_4" type="radio" /><input id="__tabbed_4_12" name="__tabbed_4" type="radio" /><input id="__tabbed_4_13" name="__tabbed_4" type="radio" /><input id="__tabbed_4_14" name="__tabbed_4" type="radio" /><div class="tabbed-labels"><label for="__tabbed_4_1">&lt;1&gt;</label><label for="__tabbed_4_2">&lt;2&gt;</label><label for="__tabbed_4_3">&lt;3&gt;</label><label for="__tabbed_4_4">&lt;4&gt;</label><label for="__tabbed_4_5">&lt;5&gt;</label><label for="__tabbed_4_6">&lt;6&gt;</label><label for="__tabbed_4_7">&lt;7&gt;</label><label for="__tabbed_4_8">&lt;8&gt;</label><label for="__tabbed_4_9">&lt;9&gt;</label><label for="__tabbed_4_10">&lt;10&gt;</label><label for="__tabbed_4_11">&lt;11&gt;</label><label for="__tabbed_4_12">&lt;12&gt;</label><label for="__tabbed_4_13">&lt;13&gt;</label><label for="__tabbed_4_14">&lt;14&gt;</label></div>
<div class="tabbed-content">
<div class="tabbed-block">
@ -2674,8 +2694,8 @@ dp[i, c] = \max(dp[i-1, c], dp[i-1, c - wgt[i-1]] + val[i-1])
</div>
</div>
</div>
<p><strong>最后考虑状态压缩</strong>。以上代码中的 <code>dp</code> 矩阵占用 <span class="arithmatex">\(O(n \times cap)\)</span> 空间。由于每个状态都只与其上一行的状态有关，因此我们可以使用两个数组滚动前进，将空间复杂度从 <span class="arithmatex">\(O(n^2)\)</span> 将低至 <span class="arithmatex">\(O(n)\)</span> 。代码省略，有兴趣的同学可以自行实现。</p>
<p>那么，我们是否可以仅用一个数组实现状态压缩呢？观察可知，每个状态都是由左上方或正上方的格子转移过来的。假设只有一个数组，当遍历到第 <span class="arithmatex">\(i\)</span> 行时，该数组存储的仍然是第 <span class="arithmatex">\(i-1\)</span> 行的状态，为了避免左边区域的格子在状态转移中被覆盖，我们应采取倒序遍历。</p>
<p><strong>最后考虑状态压缩</strong>。以上代码中的数组 <code>dp</code> 占用 <span class="arithmatex">\(O(n \times cap)\)</span> 空间。由于每个状态都只与其上一行的状态有关，因此我们可以使用两个数组滚动前进，将空间复杂度从 <span class="arithmatex">\(O(n^2)\)</span> 将低至 <span class="arithmatex">\(O(n)\)</span> 。代码省略，有兴趣的同学可以自行实现。</p>
<p>那么，我们是否可以仅用一个数组实现状态压缩呢？观察可知，每个状态都是由左上方或正上方的格子转移过来的。假设只有一个数组，当遍历到第 <span class="arithmatex">\(i\)</span> 行时，该数组存储的仍然是第 <span class="arithmatex">\(i-1\)</span> 行的状态，<strong>为了避免左方区域的格子在状态转移中被覆盖，应该采取倒序遍历</strong>。</p>
<p>以下动画展示了在单个数组下从第 <span class="arithmatex">\(i=1\)</span> 行转换至第 <span class="arithmatex">\(i=2\)</span> 行的过程。建议你思考一下正序遍历和倒序遍历的区别。</p>
<div class="tabbed-set tabbed-alternate" data-tabs="5:6"><input checked="checked" id="__tabbed_5_1" name="__tabbed_5" type="radio" /><input id="__tabbed_5_2" name="__tabbed_5" type="radio" /><input id="__tabbed_5_3" name="__tabbed_5" type="radio" /><input id="__tabbed_5_4" name="__tabbed_5" type="radio" /><input id="__tabbed_5_5" name="__tabbed_5" type="radio" /><input id="__tabbed_5_6" name="__tabbed_5" type="radio" /><div class="tabbed-labels"><label for="__tabbed_5_1">&lt;1&gt;</label><label for="__tabbed_5_2">&lt;2&gt;</label><label for="__tabbed_5_3">&lt;3&gt;</label><label for="__tabbed_5_4">&lt;4&gt;</label><label for="__tabbed_5_5">&lt;5&gt;</label><label for="__tabbed_5_6">&lt;6&gt;</label></div>
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</div>
</div>
</div>
<p>如以下代码所示，我们仅需将 <code>dp</code> 矩阵的第一维 <span class="arithmatex">\(i\)</span> 直接删除，并且将内循环修改为倒序遍历即可。</p>
<p>如以下代码所示，我们仅需将数组 <code>dp</code> 的第一维 <span class="arithmatex">\(i\)</span> 直接删除，并且将内循环修改为倒序遍历即可。</p>
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@ -2907,13 +2927,13 @@ dp[i, c] = \max(dp[i-1, c], dp[i-1, c - wgt[i-1]] + val[i-1])
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