deploy

chapter_dynamic_programming/intro_to_dynamic_programming/index.html

@@ -3414,13 +3414,13 @@
 <p>在本节中，我们从一个经典例题入手，先给出它的暴力回溯解法，观察其中包含的重叠子问题，再逐步导出更高效的动态规划解法。</p>
 <div class="admonition question">
 <p class="admonition-title">爬楼梯</p>
-<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，请问有多少种方案可以爬到楼顶。</p>
+<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，请问有多少种方案可以爬到楼顶？</p>
 </div>
 <p>如图 14-1 所示，对于一个 <span class="arithmatex">\(3\)</span> 阶楼梯，共有 <span class="arithmatex">\(3\)</span> 种方案可以爬到楼顶。</p>
 <p><a class="glightbox" href="../intro_to_dynamic_programming.assets/climbing_stairs_example.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="爬到第 3 阶的方案数量" class="animation-figure" src="../intro_to_dynamic_programming.assets/climbing_stairs_example.png" /></a></p>
 <p align="center"> 图 14-1 &nbsp; 爬到第 3 阶的方案数量 </p>
 
-<p>本题的目标是求解方案数量，<strong>我们可以考虑通过回溯来穷举所有可能性</strong>。具体来说，将爬楼梯想象为一个多轮选择的过程：从地面出发，每轮选择上 <span class="arithmatex">\(1\)</span> 阶或 <span class="arithmatex">\(2\)</span> 阶，每当到达楼梯顶部时就将方案数量加 <span class="arithmatex">\(1\)</span> ，当越过楼梯顶部时就将其剪枝。</p>
+<p>本题的目标是求解方案数量，<strong>我们可以考虑通过回溯来穷举所有可能性</strong>。具体来说，将爬楼梯想象为一个多轮选择的过程：从地面出发，每轮选择上 <span class="arithmatex">\(1\)</span> 阶或 <span class="arithmatex">\(2\)</span> 阶，每当到达楼梯顶部时就将方案数量加 <span class="arithmatex">\(1\)</span> ，当越过楼梯顶部时就将其剪枝。代码如下所示：</p>
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@@ -3440,7 +3440,7 @@
 <a id="__codelineno-0-14" name="__codelineno-0-14" href="#__codelineno-0-14"></a>
 <a id="__codelineno-0-15" name="__codelineno-0-15" href="#__codelineno-0-15"></a><span class="k">def</span> <span class="nf">climbing_stairs_backtrack</span><span class="p">(</span><span class="n">n</span><span class="p">:</span> <span class="nb">int</span><span class="p">)</span> <span class="o">-&gt;</span> <span class="nb">int</span><span class="p">:</span>
 <a id="__codelineno-0-16" name="__codelineno-0-16" href="#__codelineno-0-16"></a><span class="w">    </span><span class="sd">&quot;&quot;&quot;爬楼梯：回溯&quot;&quot;&quot;</span>
-<a id="__codelineno-0-17" name="__codelineno-0-17" href="#__codelineno-0-17"></a>    <span class="n">choices</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="p">]</span>  <span class="c1"># 可选择向上爬 1 或 2 阶</span>
+<a id="__codelineno-0-17" name="__codelineno-0-17" href="#__codelineno-0-17"></a>    <span class="n">choices</span> <span class="o">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="p">]</span>  <span class="c1"># 可选择向上爬 1 阶或 2 阶</span>
 <a id="__codelineno-0-18" name="__codelineno-0-18" href="#__codelineno-0-18"></a>    <span class="n">state</span> <span class="o">=</span> <span class="mi">0</span>  <span class="c1"># 从第 0 阶开始爬</span>
 <a id="__codelineno-0-19" name="__codelineno-0-19" href="#__codelineno-0-19"></a>    <span class="n">res</span> <span class="o">=</span> <span class="p">[</span><span class="mi">0</span><span class="p">]</span>  <span class="c1"># 使用 res[0] 记录方案数量</span>
 <a id="__codelineno-0-20" name="__codelineno-0-20" href="#__codelineno-0-20"></a>    <span class="n">backtrack</span><span class="p">(</span><span class="n">choices</span><span class="p">,</span> <span class="n">state</span><span class="p">,</span> <span class="n">n</span><span class="p">,</span> <span class="n">res</span><span class="p">)</span>
@@ -3466,7 +3466,7 @@
 <a id="__codelineno-1-16" name="__codelineno-1-16" href="#__codelineno-1-16"></a>
 <a id="__codelineno-1-17" name="__codelineno-1-17" href="#__codelineno-1-17"></a><span class="cm">/* 爬楼梯：回溯 */</span>
 <a id="__codelineno-1-18" name="__codelineno-1-18" href="#__codelineno-1-18"></a><span class="kt">int</span><span class="w"> </span><span class="nf">climbingStairsBacktrack</span><span class="p">(</span><span class="kt">int</span><span class="w"> </span><span class="n">n</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
-<a id="__codelineno-1-19" name="__codelineno-1-19" href="#__codelineno-1-19"></a><span class="w">    </span><span class="n">vector</span><span class="o">&lt;</span><span class="kt">int</span><span class="o">&gt;</span><span class="w"> </span><span class="n">choices</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="p">{</span><span class="mi">1</span><span class="p">,</span><span class="w"> </span><span class="mi">2</span><span class="p">};</span><span class="w"> </span><span class="c1">// 可选择向上爬 1 或 2 阶</span>
+<a id="__codelineno-1-19" name="__codelineno-1-19" href="#__codelineno-1-19"></a><span class="w">    </span><span class="n">vector</span><span class="o">&lt;</span><span class="kt">int</span><span class="o">&gt;</span><span class="w"> </span><span class="n">choices</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="p">{</span><span class="mi">1</span><span class="p">,</span><span class="w"> </span><span class="mi">2</span><span class="p">};</span><span class="w"> </span><span class="c1">// 可选择向上爬 1 阶或 2 阶</span>
 <a id="__codelineno-1-20" name="__codelineno-1-20" href="#__codelineno-1-20"></a><span class="w">    </span><span class="kt">int</span><span class="w"> </span><span class="n">state</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="mi">0</span><span class="p">;</span><span class="w">                </span><span class="c1">// 从第 0 阶开始爬</span>
 <a id="__codelineno-1-21" name="__codelineno-1-21" href="#__codelineno-1-21"></a><span class="w">    </span><span class="n">vector</span><span class="o">&lt;</span><span class="kt">int</span><span class="o">&gt;</span><span class="w"> </span><span class="n">res</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="p">{</span><span class="mi">0</span><span class="p">};</span><span class="w">        </span><span class="c1">// 使用 res[0] 记录方案数量</span>
 <a id="__codelineno-1-22" name="__codelineno-1-22" href="#__codelineno-1-22"></a><span class="w">    </span><span class="n">backtrack</span><span class="p">(</span><span class="n">choices</span><span class="p">,</span><span class="w"> </span><span class="n">state</span><span class="p">,</span><span class="w"> </span><span class="n">n</span><span class="p">,</span><span class="w"> </span><span class="n">res</span><span class="p">);</span>
@@ -3493,7 +3493,7 @@
 <a id="__codelineno-2-16" name="__codelineno-2-16" href="#__codelineno-2-16"></a>
 <a id="__codelineno-2-17" name="__codelineno-2-17" href="#__codelineno-2-17"></a><span class="cm">/* 爬楼梯：回溯 */</span>
 <a id="__codelineno-2-18" name="__codelineno-2-18" href="#__codelineno-2-18"></a><span class="kt">int</span><span class="w"> </span><span class="nf">climbingStairsBacktrack</span><span class="p">(</span><span class="kt">int</span><span class="w"> </span><span class="n">n</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
-<a id="__codelineno-2-19" name="__codelineno-2-19" href="#__codelineno-2-19"></a><span class="w">    </span><span class="n">List</span><span class="o">&lt;</span><span class="n">Integer</span><span class="o">&gt;</span><span class="w"> </span><span class="n">choices</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">Arrays</span><span class="p">.</span><span class="na">asList</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="w"> </span><span class="mi">2</span><span class="p">);</span><span class="w"> </span><span class="c1">// 可选择向上爬 1 或 2 阶</span>
+<a id="__codelineno-2-19" name="__codelineno-2-19" href="#__codelineno-2-19"></a><span class="w">    </span><span class="n">List</span><span class="o">&lt;</span><span class="n">Integer</span><span class="o">&gt;</span><span class="w"> </span><span class="n">choices</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">Arrays</span><span class="p">.</span><span class="na">asList</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span><span class="w"> </span><span class="mi">2</span><span class="p">);</span><span class="w"> </span><span class="c1">// 可选择向上爬 1 阶或 2 阶</span>
 <a id="__codelineno-2-20" name="__codelineno-2-20" href="#__codelineno-2-20"></a><span class="w">    </span><span class="kt">int</span><span class="w"> </span><span class="n">state</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="mi">0</span><span class="p">;</span><span class="w"> </span><span class="c1">// 从第 0 阶开始爬</span>
 <a id="__codelineno-2-21" name="__codelineno-2-21" href="#__codelineno-2-21"></a><span class="w">    </span><span class="n">List</span><span class="o">&lt;</span><span class="n">Integer</span><span class="o">&gt;</span><span class="w"> </span><span class="n">res</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="k">new</span><span class="w"> </span><span class="n">ArrayList</span><span class="o">&lt;&gt;</span><span class="p">();</span>
 <a id="__codelineno-2-22" name="__codelineno-2-22" href="#__codelineno-2-22"></a><span class="w">    </span><span class="n">res</span><span class="p">.</span><span class="na">add</span><span class="p">(</span><span class="mi">0</span><span class="p">);</span><span class="w"> </span><span class="c1">// 使用 res[0] 记录方案数量</span>
@@ -3521,7 +3521,7 @@
 <a id="__codelineno-3-16" name="__codelineno-3-16" href="#__codelineno-3-16"></a>
 <a id="__codelineno-3-17" name="__codelineno-3-17" href="#__codelineno-3-17"></a><span class="cm">/* 爬楼梯：回溯 */</span>
 <a id="__codelineno-3-18" name="__codelineno-3-18" href="#__codelineno-3-18"></a><span class="kt">int</span><span class="w"> </span><span class="nf">ClimbingStairsBacktrack</span><span class="p">(</span><span class="kt">int</span><span class="w"> </span><span class="n">n</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
-<a id="__codelineno-3-19" name="__codelineno-3-19" href="#__codelineno-3-19"></a><span class="w">    </span><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">choices</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="p">[</span><span class="m">1</span><span class="p">,</span><span class="w"> </span><span class="m">2</span><span class="p">];</span><span class="w"> </span><span class="c1">// 可选择向上爬 1 或 2 阶</span>
+<a id="__codelineno-3-19" name="__codelineno-3-19" href="#__codelineno-3-19"></a><span class="w">    </span><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">choices</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="p">[</span><span class="m">1</span><span class="p">,</span><span class="w"> </span><span class="m">2</span><span class="p">];</span><span class="w"> </span><span class="c1">// 可选择向上爬 1 阶或 2 阶</span>
 <a id="__codelineno-3-20" name="__codelineno-3-20" href="#__codelineno-3-20"></a><span class="w">    </span><span class="kt">int</span><span class="w"> </span><span class="n">state</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="m">0</span><span class="p">;</span><span class="w"> </span><span class="c1">// 从第 0 阶开始爬</span>
 <a id="__codelineno-3-21" name="__codelineno-3-21" href="#__codelineno-3-21"></a><span class="w">    </span><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">res</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="p">[</span><span class="m">0</span><span class="p">];</span><span class="w"> </span><span class="c1">// 使用 res[0] 记录方案数量</span>
 <a id="__codelineno-3-22" name="__codelineno-3-22" href="#__codelineno-3-22"></a><span class="w">    </span><span class="n">Backtrack</span><span class="p">(</span><span class="n">choices</span><span class="p">,</span><span class="w"> </span><span class="n">state</span><span class="p">,</span><span class="w"> </span><span class="n">n</span><span class="p">,</span><span class="w"> </span><span class="n">res</span><span class="p">);</span>
@@ -3550,7 +3550,7 @@
 <a id="__codelineno-4-18" name="__codelineno-4-18" href="#__codelineno-4-18"></a>
 <a id="__codelineno-4-19" name="__codelineno-4-19" href="#__codelineno-4-19"></a><span class="cm">/* 爬楼梯：回溯 */</span>
 <a id="__codelineno-4-20" name="__codelineno-4-20" href="#__codelineno-4-20"></a><span class="kd">func</span><span class="w"> </span><span class="nx">climbingStairsBacktrack</span><span class="p">(</span><span class="nx">n</span><span class="w"> </span><span class="kt">int</span><span class="p">)</span><span class="w"> </span><span class="kt">int</span><span class="w"> </span><span class="p">{</span>
-<a id="__codelineno-4-21" name="__codelineno-4-21" href="#__codelineno-4-21"></a><span class="w">    </span><span class="c1">// 可选择向上爬 1 或 2 阶</span>
+<a id="__codelineno-4-21" name="__codelineno-4-21" href="#__codelineno-4-21"></a><span class="w">    </span><span class="c1">// 可选择向上爬 1 阶或 2 阶</span>
 <a id="__codelineno-4-22" name="__codelineno-4-22" href="#__codelineno-4-22"></a><span class="w">    </span><span class="nx">choices</span><span class="w"> </span><span class="o">:=</span><span class="w"> </span><span class="p">[]</span><span class="kt">int</span><span class="p">{</span><span class="mi">1</span><span class="p">,</span><span class="w"> </span><span class="mi">2</span><span class="p">}</span>
 <a id="__codelineno-4-23" name="__codelineno-4-23" href="#__codelineno-4-23"></a><span class="w">    </span><span class="c1">// 从第 0 阶开始爬</span>
 <a id="__codelineno-4-24" name="__codelineno-4-24" href="#__codelineno-4-24"></a><span class="w">    </span><span class="nx">state</span><span class="w"> </span><span class="o">:=</span><span class="w"> </span><span class="mi">0</span>
@@ -3581,7 +3581,7 @@
 <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="cm">/* 爬楼梯：回溯 */</span>
 <a id="__codelineno-5-18" name="__codelineno-5-18" href="#__codelineno-5-18"></a><span class="kd">func</span> <span class="nf">climbingStairsBacktrack</span><span class="p">(</span><span class="n">n</span><span class="p">:</span> <span class="nb">Int</span><span class="p">)</span> <span class="p">-&gt;</span> <span class="nb">Int</span> <span class="p">{</span>
-<a id="__codelineno-5-19" name="__codelineno-5-19" href="#__codelineno-5-19"></a>    <span class="kd">let</span> <span class="nv">choices</span> <span class="p">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="p">]</span> <span class="c1">// 可选择向上爬 1 或 2 阶</span>
+<a id="__codelineno-5-19" name="__codelineno-5-19" href="#__codelineno-5-19"></a>    <span class="kd">let</span> <span class="nv">choices</span> <span class="p">=</span> <span class="p">[</span><span class="mi">1</span><span class="p">,</span> <span class="mi">2</span><span class="p">]</span> <span class="c1">// 可选择向上爬 1 阶或 2 阶</span>
 <a id="__codelineno-5-20" name="__codelineno-5-20" href="#__codelineno-5-20"></a>    <span class="kd">let</span> <span class="nv">state</span> <span class="p">=</span> <span class="mi">0</span> <span class="c1">// 从第 0 阶开始爬</span>
 <a id="__codelineno-5-21" name="__codelineno-5-21" href="#__codelineno-5-21"></a>    <span class="kd">var</span> <span class="nv">res</span><span class="p">:</span> <span class="p">[</span><span class="nb">Int</span><span class="p">]</span> <span class="p">=</span> <span class="p">[]</span>
 <a id="__codelineno-5-22" name="__codelineno-5-22" href="#__codelineno-5-22"></a>    <span class="n">res</span><span class="p">.</span><span class="n">append</span><span class="p">(</span><span class="mi">0</span><span class="p">)</span> <span class="c1">// 使用 res[0] 记录方案数量</span>
@@ -3607,7 +3607,7 @@
 <a id="__codelineno-6-14" name="__codelineno-6-14" href="#__codelineno-6-14"></a>
 <a id="__codelineno-6-15" name="__codelineno-6-15" href="#__codelineno-6-15"></a><span class="cm">/* 爬楼梯：回溯 */</span>
 <a id="__codelineno-6-16" name="__codelineno-6-16" href="#__codelineno-6-16"></a><span class="kd">function</span><span class="w"> </span><span class="nx">climbingStairsBacktrack</span><span class="p">(</span><span class="nx">n</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
-<a id="__codelineno-6-17" name="__codelineno-6-17" href="#__codelineno-6-17"></a><span class="w">    </span><span class="kd">const</span><span class="w"> </span><span class="nx">choices</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="p">[</span><span class="mf">1</span><span class="p">,</span><span class="w"> </span><span class="mf">2</span><span class="p">];</span><span class="w"> </span><span class="c1">// 可选择向上爬 1 或 2 阶</span>
+<a id="__codelineno-6-17" name="__codelineno-6-17" href="#__codelineno-6-17"></a><span class="w">    </span><span class="kd">const</span><span class="w"> </span><span class="nx">choices</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="p">[</span><span class="mf">1</span><span class="p">,</span><span class="w"> </span><span class="mf">2</span><span class="p">];</span><span class="w"> </span><span class="c1">// 可选择向上爬 1 阶或 2 阶</span>
 <a id="__codelineno-6-18" name="__codelineno-6-18" href="#__codelineno-6-18"></a><span class="w">    </span><span class="kd">const</span><span class="w"> </span><span class="nx">state</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="mf">0</span><span class="p">;</span><span class="w"> </span><span class="c1">// 从第 0 阶开始爬</span>
 <a id="__codelineno-6-19" name="__codelineno-6-19" href="#__codelineno-6-19"></a><span class="w">    </span><span class="kd">const</span><span class="w"> </span><span class="nx">res</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="ow">new</span><span class="w"> </span><span class="nb">Map</span><span class="p">();</span>
 <a id="__codelineno-6-20" name="__codelineno-6-20" href="#__codelineno-6-20"></a><span class="w">    </span><span class="nx">res</span><span class="p">.</span><span class="nx">set</span><span class="p">(</span><span class="mf">0</span><span class="p">,</span><span class="w"> </span><span class="mf">0</span><span class="p">);</span><span class="w"> </span><span class="c1">// 使用 res[0] 记录方案数量</span>
@@ -3638,7 +3638,7 @@
 <a id="__codelineno-7-19" name="__codelineno-7-19" href="#__codelineno-7-19"></a>
 <a id="__codelineno-7-20" name="__codelineno-7-20" href="#__codelineno-7-20"></a><span class="cm">/* 爬楼梯：回溯 */</span>
 <a id="__codelineno-7-21" name="__codelineno-7-21" href="#__codelineno-7-21"></a><span class="kd">function</span><span class="w"> </span><span class="nx">climbingStairsBacktrack</span><span class="p">(</span><span class="nx">n</span><span class="o">:</span><span class="w"> </span><span class="kt">number</span><span class="p">)</span><span class="o">:</span><span class="w"> </span><span class="kt">number</span><span class="w"> </span><span class="p">{</span>
-<a id="__codelineno-7-22" name="__codelineno-7-22" href="#__codelineno-7-22"></a><span class="w">    </span><span class="kd">const</span><span class="w"> </span><span class="nx">choices</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="p">[</span><span class="mf">1</span><span class="p">,</span><span class="w"> </span><span class="mf">2</span><span class="p">];</span><span class="w"> </span><span class="c1">// 可选择向上爬 1 或 2 阶</span>
+<a id="__codelineno-7-22" name="__codelineno-7-22" href="#__codelineno-7-22"></a><span class="w">    </span><span class="kd">const</span><span class="w"> </span><span class="nx">choices</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="p">[</span><span class="mf">1</span><span class="p">,</span><span class="w"> </span><span class="mf">2</span><span class="p">];</span><span class="w"> </span><span class="c1">// 可选择向上爬 1 阶或 2 阶</span>
 <a id="__codelineno-7-23" name="__codelineno-7-23" href="#__codelineno-7-23"></a><span class="w">    </span><span class="kd">const</span><span class="w"> </span><span class="nx">state</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="mf">0</span><span class="p">;</span><span class="w"> </span><span class="c1">// 从第 0 阶开始爬</span>
 <a id="__codelineno-7-24" name="__codelineno-7-24" href="#__codelineno-7-24"></a><span class="w">    </span><span class="kd">const</span><span class="w"> </span><span class="nx">res</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="ow">new</span><span class="w"> </span><span class="nb">Map</span><span class="p">();</span>
 <a id="__codelineno-7-25" name="__codelineno-7-25" href="#__codelineno-7-25"></a><span class="w">    </span><span class="nx">res</span><span class="p">.</span><span class="nx">set</span><span class="p">(</span><span class="mf">0</span><span class="p">,</span><span class="w"> </span><span class="mf">0</span><span class="p">);</span><span class="w"> </span><span class="c1">// 使用 res[0] 记录方案数量</span>
@@ -3666,7 +3666,7 @@
 <a id="__codelineno-8-16" name="__codelineno-8-16" href="#__codelineno-8-16"></a>
 <a id="__codelineno-8-17" name="__codelineno-8-17" href="#__codelineno-8-17"></a><span class="cm">/* 爬楼梯：回溯 */</span>
 <a id="__codelineno-8-18" name="__codelineno-8-18" href="#__codelineno-8-18"></a><span class="kt">int</span><span class="w"> </span><span class="n">climbingStairsBacktrack</span><span class="p">(</span><span class="kt">int</span><span class="w"> </span><span class="n">n</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
-<a id="__codelineno-8-19" name="__codelineno-8-19" href="#__codelineno-8-19"></a><span class="w">  </span><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">choices</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="p">[</span><span class="m">1</span><span class="p">,</span><span class="w"> </span><span class="m">2</span><span class="p">];</span><span class="w"> </span><span class="c1">// 可选择向上爬 1 或 2 阶</span>
+<a id="__codelineno-8-19" name="__codelineno-8-19" href="#__codelineno-8-19"></a><span class="w">  </span><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">choices</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="p">[</span><span class="m">1</span><span class="p">,</span><span class="w"> </span><span class="m">2</span><span class="p">];</span><span class="w"> </span><span class="c1">// 可选择向上爬 1 阶或 2 阶</span>
 <a id="__codelineno-8-20" name="__codelineno-8-20" href="#__codelineno-8-20"></a><span class="w">  </span><span class="kt">int</span><span class="w"> </span><span class="n">state</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="m">0</span><span class="p">;</span><span class="w"> </span><span class="c1">// 从第 0 阶开始爬</span>
 <a id="__codelineno-8-21" name="__codelineno-8-21" href="#__codelineno-8-21"></a><span class="w">  </span><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">res</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="p">[];</span>
 <a id="__codelineno-8-22" name="__codelineno-8-22" href="#__codelineno-8-22"></a><span class="w">  </span><span class="n">res</span><span class="p">.</span><span class="n">add</span><span class="p">(</span><span class="m">0</span><span class="p">);</span><span class="w"> </span><span class="c1">// 使用 res[0] 记录方案数量</span>
@@ -3692,7 +3692,7 @@
 <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="cm">/* 爬楼梯：回溯 */</span>
 <a id="__codelineno-9-16" name="__codelineno-9-16" href="#__codelineno-9-16"></a><span class="k">fn</span> <span class="nf">climbing_stairs_backtrack</span><span class="p">(</span><span class="n">n</span>: <span class="kt">usize</span><span class="p">)</span><span class="w"> </span>-&gt; <span class="kt">i32</span> <span class="p">{</span>
-<a id="__codelineno-9-17" name="__codelineno-9-17" href="#__codelineno-9-17"></a><span class="w">    </span><span class="kd">let</span><span class="w"> </span><span class="n">choices</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="fm">vec!</span><span class="p">[</span><span class="w"> </span><span class="mi">1</span><span class="p">,</span><span class="w"> </span><span class="mi">2</span><span class="w"> </span><span class="p">];</span><span class="w"> </span><span class="c1">// 可选择向上爬 1 或 2 阶</span>
+<a id="__codelineno-9-17" name="__codelineno-9-17" href="#__codelineno-9-17"></a><span class="w">    </span><span class="kd">let</span><span class="w"> </span><span class="n">choices</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="fm">vec!</span><span class="p">[</span><span class="w"> </span><span class="mi">1</span><span class="p">,</span><span class="w"> </span><span class="mi">2</span><span class="w"> </span><span class="p">];</span><span class="w"> </span><span class="c1">// 可选择向上爬 1 阶或 2 阶</span>
 <a id="__codelineno-9-18" name="__codelineno-9-18" href="#__codelineno-9-18"></a><span class="w">    </span><span class="kd">let</span><span class="w"> </span><span class="n">state</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="mi">0</span><span class="p">;</span><span class="w"> </span><span class="c1">// 从第 0 阶开始爬</span>
 <a id="__codelineno-9-19" name="__codelineno-9-19" href="#__codelineno-9-19"></a><span class="w">    </span><span class="kd">let</span><span class="w"> </span><span class="k">mut</span><span class="w"> </span><span class="n">res</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="nb">Vec</span>::<span class="n">new</span><span class="p">();</span>
 <a id="__codelineno-9-20" name="__codelineno-9-20" href="#__codelineno-9-20"></a><span class="w">    </span><span class="n">res</span><span class="p">.</span><span class="n">push</span><span class="p">(</span><span class="mi">0</span><span class="p">);</span><span class="w"> </span><span class="c1">// 使用 res[0] 记录方案数量</span>
@@ -3721,7 +3721,7 @@
 <a id="__codelineno-10-17" name="__codelineno-10-17" href="#__codelineno-10-17"></a>
 <a id="__codelineno-10-18" name="__codelineno-10-18" href="#__codelineno-10-18"></a><span class="cm">/* 爬楼梯：回溯 */</span>
 <a id="__codelineno-10-19" name="__codelineno-10-19" href="#__codelineno-10-19"></a><span class="kt">int</span><span class="w"> </span><span class="nf">climbingStairsBacktrack</span><span class="p">(</span><span class="kt">int</span><span class="w"> </span><span class="n">n</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
-<a id="__codelineno-10-20" name="__codelineno-10-20" href="#__codelineno-10-20"></a><span class="w">    </span><span class="kt">int</span><span class="w"> </span><span class="n">choices</span><span class="p">[</span><span class="mi">2</span><span class="p">]</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="p">{</span><span class="mi">1</span><span class="p">,</span><span class="w"> </span><span class="mi">2</span><span class="p">};</span><span class="w"> </span><span class="c1">// 可选择向上爬 1 或 2 阶</span>
+<a id="__codelineno-10-20" name="__codelineno-10-20" href="#__codelineno-10-20"></a><span class="w">    </span><span class="kt">int</span><span class="w"> </span><span class="n">choices</span><span class="p">[</span><span class="mi">2</span><span class="p">]</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="p">{</span><span class="mi">1</span><span class="p">,</span><span class="w"> </span><span class="mi">2</span><span class="p">};</span><span class="w"> </span><span class="c1">// 可选择向上爬 1 阶或 2 阶</span>
 <a id="__codelineno-10-21" name="__codelineno-10-21" href="#__codelineno-10-21"></a><span class="w">    </span><span class="kt">int</span><span class="w"> </span><span class="n">state</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="mi">0</span><span class="p">;</span><span class="w">           </span><span class="c1">// 从第 0 阶开始爬</span>
 <a id="__codelineno-10-22" name="__codelineno-10-22" href="#__codelineno-10-22"></a><span class="w">    </span><span class="kt">int</span><span class="w"> </span><span class="o">*</span><span class="n">res</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="p">(</span><span class="kt">int</span><span class="w"> </span><span class="o">*</span><span class="p">)</span><span class="n">malloc</span><span class="p">(</span><span class="k">sizeof</span><span class="p">(</span><span class="kt">int</span><span class="p">));</span>
 <a id="__codelineno-10-23" name="__codelineno-10-23" href="#__codelineno-10-23"></a><span class="w">    </span><span class="o">*</span><span class="n">res</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="mi">0</span><span class="p">;</span><span class="w"> </span><span class="c1">// 使用 res[0] 记录方案数量</span>
@@ -3754,7 +3754,7 @@
 <a id="__codelineno-11-18" name="__codelineno-11-18" href="#__codelineno-11-18"></a>
 <a id="__codelineno-11-19" name="__codelineno-11-19" href="#__codelineno-11-19"></a><span class="c1">// 爬楼梯：回溯</span>
 <a id="__codelineno-11-20" name="__codelineno-11-20" href="#__codelineno-11-20"></a><span class="k">fn</span><span class="w"> </span><span class="n">climbingStairsBacktrack</span><span class="p">(</span><span class="n">n</span><span class="o">:</span><span class="w"> </span><span class="kt">usize</span><span class="p">)</span><span class="w"> </span><span class="o">!</span><span class="kt">i32</span><span class="w"> </span><span class="p">{</span>
-<a id="__codelineno-11-21" name="__codelineno-11-21" href="#__codelineno-11-21"></a><span class="w">    </span><span class="kr">var</span><span class="w"> </span><span class="n">choices</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="p">[</span><span class="n">_</span><span class="p">]</span><span class="kt">i32</span><span class="p">{</span><span class="w"> </span><span class="mi">1</span><span class="p">,</span><span class="w"> </span><span class="mi">2</span><span class="w"> </span><span class="p">};</span><span class="w"> </span><span class="c1">// 可选择向上爬 1 或 2 阶</span>
+<a id="__codelineno-11-21" name="__codelineno-11-21" href="#__codelineno-11-21"></a><span class="w">    </span><span class="kr">var</span><span class="w"> </span><span class="n">choices</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="p">[</span><span class="n">_</span><span class="p">]</span><span class="kt">i32</span><span class="p">{</span><span class="w"> </span><span class="mi">1</span><span class="p">,</span><span class="w"> </span><span class="mi">2</span><span class="w"> </span><span class="p">};</span><span class="w"> </span><span class="c1">// 可选择向上爬 1 阶或 2 阶</span>
 <a id="__codelineno-11-22" name="__codelineno-11-22" href="#__codelineno-11-22"></a><span class="w">    </span><span class="kr">var</span><span class="w"> </span><span class="n">state</span><span class="o">:</span><span class="w"> </span><span class="kt">i32</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="mi">0</span><span class="p">;</span><span class="w"> </span><span class="c1">// 从第 0 阶开始爬</span>
 <a id="__codelineno-11-23" name="__codelineno-11-23" href="#__codelineno-11-23"></a><span class="w">    </span><span class="kr">var</span><span class="w"> </span><span class="n">res</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">std</span><span class="p">.</span><span class="n">ArrayList</span><span class="p">(</span><span class="kt">i32</span><span class="p">).</span><span class="n">init</span><span class="p">(</span><span class="n">std</span><span class="p">.</span><span class="n">heap</span><span class="p">.</span><span class="n">page_allocator</span><span class="p">);</span>
 <a id="__codelineno-11-24" name="__codelineno-11-24" href="#__codelineno-11-24"></a><span class="w">    </span><span class="k">defer</span><span class="w"> </span><span class="n">res</span><span class="p">.</span><span class="n">deinit</span><span class="p">();</span>
@@ -3767,12 +3767,12 @@
 </div>
 </div>
 <h2 id="1411">14.1.1 &nbsp; 方法一：暴力搜索<a class="headerlink" href="#1411" title="Permanent link">&para;</a></h2>
-<p>回溯算法通常并不显式地对问题进行拆解，而是将问题看作一系列决策步骤，通过试探和剪枝，搜索所有可能的解。</p>
-<p>我们可以尝试从问题分解的角度分析这道题。设爬到第 <span class="arithmatex">\(i\)</span> 阶共有 <span class="arithmatex">\(dp[i]\)</span> 种方案，那么 <span class="arithmatex">\(dp[i]\)</span> 就是原问题，其子问题包括:</p>
+<p>回溯算法通常并不显式地对问题进行拆解，而是将求解问题看作一系列决策步骤，通过试探和剪枝，搜索所有可能的解。</p>
+<p>我们可以尝试从问题分解的角度分析这道题。设爬到第 <span class="arithmatex">\(i\)</span> 阶共有 <span class="arithmatex">\(dp[i]\)</span> 种方案，那么 <span class="arithmatex">\(dp[i]\)</span> 就是原问题，其子问题包括：</p>
 <div class="arithmatex">\[
 dp[i-1], dp[i-2], \dots, dp[2], dp[1]
 \]</div>
-<p>由于每轮只能上 <span class="arithmatex">\(1\)</span> 阶或 <span class="arithmatex">\(2\)</span> 阶，因此当我们站在第 <span class="arithmatex">\(i\)</span> 阶楼梯上时，上一轮只可能站在第 <span class="arithmatex">\(i - 1\)</span> 阶或第 <span class="arithmatex">\(i - 2\)</span> 阶上。换句话说，我们只能从第 <span class="arithmatex">\(i -1\)</span> 阶或第 <span class="arithmatex">\(i - 2\)</span> 阶前往第 <span class="arithmatex">\(i\)</span> 阶。</p>
+<p>由于每轮只能上 <span class="arithmatex">\(1\)</span> 阶或 <span class="arithmatex">\(2\)</span> 阶，因此当我们站在第 <span class="arithmatex">\(i\)</span> 阶楼梯上时，上一轮只可能站在第 <span class="arithmatex">\(i - 1\)</span> 阶或第 <span class="arithmatex">\(i - 2\)</span> 阶上。换句话说，我们只能从第 <span class="arithmatex">\(i -1\)</span> 阶或第 <span class="arithmatex">\(i - 2\)</span> 阶迈向第 <span class="arithmatex">\(i\)</span> 阶。</p>
 <p>由此便可得出一个重要推论：<strong>爬到第 <span class="arithmatex">\(i - 1\)</span> 阶的方案数加上爬到第 <span class="arithmatex">\(i - 2\)</span> 阶的方案数就等于爬到第 <span class="arithmatex">\(i\)</span> 阶的方案数</strong>。公式如下：</p>
 <div class="arithmatex">\[
 dp[i] = dp[i-1] + dp[i-2]
@@ -3782,7 +3782,7 @@ dp[i] = dp[i-1] + dp[i-2]
 <p align="center"> 图 14-2 &nbsp; 方案数量递推关系 </p>
 
 <p>我们可以根据递推公式得到暴力搜索解法。以 <span class="arithmatex">\(dp[n]\)</span> 为起始点，<strong>递归地将一个较大问题拆解为两个较小问题的和</strong>，直至到达最小子问题 <span class="arithmatex">\(dp[1]\)</span> 和 <span class="arithmatex">\(dp[2]\)</span> 时返回。其中，最小子问题的解是已知的，即 <span class="arithmatex">\(dp[1] = 1\)</span>、<span class="arithmatex">\(dp[2] = 2\)</span> ，表示爬到第 <span class="arithmatex">\(1\)</span>、<span class="arithmatex">\(2\)</span> 阶分别有 <span class="arithmatex">\(1\)</span>、<span class="arithmatex">\(2\)</span> 种方案。</p>
-<p>观察以下代码，它和标准回溯代码都属于深度优先搜索，但更加简洁。</p>
+<p>观察以下代码，它和标准回溯代码都属于深度优先搜索，但更加简洁：</p>
 <div class="tabbed-set tabbed-alternate" data-tabs="2:12"><input checked="checked" id="__tabbed_2_1" name="__tabbed_2" type="radio" /><input id="__tabbed_2_2" name="__tabbed_2" type="radio" /><input id="__tabbed_2_3" name="__tabbed_2" type="radio" /><input id="__tabbed_2_4" name="__tabbed_2" type="radio" /><input id="__tabbed_2_5" name="__tabbed_2" type="radio" /><input id="__tabbed_2_6" name="__tabbed_2" type="radio" /><input id="__tabbed_2_7" name="__tabbed_2" type="radio" /><input id="__tabbed_2_8" name="__tabbed_2" type="radio" /><input id="__tabbed_2_9" name="__tabbed_2" type="radio" /><input id="__tabbed_2_10" name="__tabbed_2" type="radio" /><input id="__tabbed_2_11" name="__tabbed_2" type="radio" /><input id="__tabbed_2_12" name="__tabbed_2" type="radio" /><div class="tabbed-labels"><label for="__tabbed_2_1">Python</label><label for="__tabbed_2_2">C++</label><label for="__tabbed_2_3">Java</label><label for="__tabbed_2_4">C#</label><label for="__tabbed_2_5">Go</label><label for="__tabbed_2_6">Swift</label><label for="__tabbed_2_7">JS</label><label for="__tabbed_2_8">TS</label><label for="__tabbed_2_9">Dart</label><label for="__tabbed_2_10">Rust</label><label for="__tabbed_2_11">C</label><label for="__tabbed_2_12">Zig</label></div>
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@@ -3992,7 +3992,7 @@ dp[i] = dp[i-1] + dp[i-2]
 <p><a class="glightbox" href="../intro_to_dynamic_programming.assets/climbing_stairs_dfs_tree.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="爬楼梯对应递归树" class="animation-figure" src="../intro_to_dynamic_programming.assets/climbing_stairs_dfs_tree.png" /></a></p>
 <p align="center"> 图 14-3 &nbsp; 爬楼梯对应递归树 </p>
 
-<p>观察图 14-3 ，<strong>指数阶的时间复杂度是由于“重叠子问题”导致的</strong>。例如 <span class="arithmatex">\(dp[9]\)</span> 被分解为 <span class="arithmatex">\(dp[8]\)</span> 和 <span class="arithmatex">\(dp[7]\)</span> ，<span class="arithmatex">\(dp[8]\)</span> 被分解为 <span class="arithmatex">\(dp[7]\)</span> 和 <span class="arithmatex">\(dp[6]\)</span> ，两者都包含子问题 <span class="arithmatex">\(dp[7]\)</span> 。</p>
+<p>观察图 14-3 ，<strong>指数阶的时间复杂度是“重叠子问题”导致的</strong>。例如 <span class="arithmatex">\(dp[9]\)</span> 被分解为 <span class="arithmatex">\(dp[8]\)</span> 和 <span class="arithmatex">\(dp[7]\)</span> ，<span class="arithmatex">\(dp[8]\)</span> 被分解为 <span class="arithmatex">\(dp[7]\)</span> 和 <span class="arithmatex">\(dp[6]\)</span> ，两者都包含子问题 <span class="arithmatex">\(dp[7]\)</span> 。</p>
 <p>以此类推，子问题中包含更小的重叠子问题，子子孙孙无穷尽也。绝大部分计算资源都浪费在这些重叠的问题上。</p>
 <h2 id="1412">14.1.2 &nbsp; 方法二：记忆化搜索<a class="headerlink" href="#1412" title="Permanent link">&para;</a></h2>
 <p>为了提升算法效率，<strong>我们希望所有的重叠子问题都只被计算一次</strong>。为此，我们声明一个数组 <code>mem</code> 来记录每个子问题的解，并在搜索过程中将重叠子问题剪枝。</p>
@@ -4000,6 +4000,7 @@ dp[i] = dp[i-1] + dp[i-2]
 <li>当首次计算 <span class="arithmatex">\(dp[i]\)</span> 时，我们将其记录至 <code>mem[i]</code> ，以便之后使用。</li>
 <li>当再次需要计算 <span class="arithmatex">\(dp[i]\)</span> 时，我们便可直接从 <code>mem[i]</code> 中获取结果，从而避免重复计算该子问题。</li>
 </ol>
+<p>代码如下所示：</p>
 <div class="tabbed-set tabbed-alternate" data-tabs="3:12"><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" /><input id="__tabbed_3_12" name="__tabbed_3" type="radio" /><div class="tabbed-labels"><label for="__tabbed_3_1">Python</label><label for="__tabbed_3_2">C++</label><label for="__tabbed_3_3">Java</label><label for="__tabbed_3_4">C#</label><label for="__tabbed_3_5">Go</label><label for="__tabbed_3_6">Swift</label><label for="__tabbed_3_7">JS</label><label for="__tabbed_3_8">TS</label><label for="__tabbed_3_9">Dart</label><label for="__tabbed_3_10">Rust</label><label for="__tabbed_3_11">C</label><label for="__tabbed_3_12">Zig</label></div>
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@@ -4298,14 +4299,14 @@ dp[i] = dp[i-1] + dp[i-2]
 </div>
 </div>
 </div>
-<p>观察图 14-4 ，<strong>经过记忆化处理后，所有重叠子问题都只需被计算一次，时间复杂度被优化至 <span class="arithmatex">\(O(n)\)</span></strong> ，这是一个巨大的飞跃。</p>
+<p>观察图 14-4 ，<strong>经过记忆化处理后，所有重叠子问题都只需计算一次，时间复杂度优化至 <span class="arithmatex">\(O(n)\)</span></strong> ，这是一个巨大的飞跃。</p>
 <p><a class="glightbox" href="../intro_to_dynamic_programming.assets/climbing_stairs_dfs_memo_tree.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="记忆化搜索对应递归树" class="animation-figure" src="../intro_to_dynamic_programming.assets/climbing_stairs_dfs_memo_tree.png" /></a></p>
 <p align="center"> 图 14-4 &nbsp; 记忆化搜索对应递归树 </p>
 
 <h2 id="1413">14.1.3 &nbsp; 方法三：动态规划<a class="headerlink" href="#1413" title="Permanent link">&para;</a></h2>
-<p><strong>记忆化搜索是一种“从顶至底”的方法</strong>：我们从原问题（根节点）开始，递归地将较大子问题分解为较小子问题，直至解已知的最小子问题（叶节点）。之后，通过回溯将子问题的解逐层收集，构建出原问题的解。</p>
+<p><strong>记忆化搜索是一种“从顶至底”的方法</strong>：我们从原问题（根节点）开始，递归地将较大子问题分解为较小子问题，直至解已知的最小子问题（叶节点）。之后，通过回溯逐层收集子问题的解，构建出原问题的解。</p>
 <p>与之相反，<strong>动态规划是一种“从底至顶”的方法</strong>：从最小子问题的解开始，迭代地构建更大子问题的解，直至得到原问题的解。</p>
-<p>由于动态规划不包含回溯过程，因此只需使用循环迭代实现，无须使用递归。在以下代码中，我们初始化一个数组 <code>dp</code> 来存储子问题的解，它起到了记忆化搜索中数组 <code>mem</code> 相同的记录作用。</p>
+<p>由于动态规划不包含回溯过程，因此只需使用循环迭代实现，无须使用递归。在以下代码中，我们初始化一个数组 <code>dp</code> 来存储子问题的解，它起到了与记忆化搜索中数组 <code>mem</code> 相同的记录作用：</p>
 <div class="tabbed-set tabbed-alternate" data-tabs="4:12"><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" /><div class="tabbed-labels"><label for="__tabbed_4_1">Python</label><label for="__tabbed_4_2">C++</label><label for="__tabbed_4_3">Java</label><label for="__tabbed_4_4">C#</label><label for="__tabbed_4_5">Go</label><label for="__tabbed_4_6">Swift</label><label for="__tabbed_4_7">JS</label><label for="__tabbed_4_8">TS</label><label for="__tabbed_4_9">Dart</label><label for="__tabbed_4_10">Rust</label><label for="__tabbed_4_11">C</label><label for="__tabbed_4_12">Zig</label></div>
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@@ -4530,15 +4531,15 @@ dp[i] = dp[i-1] + dp[i-2]
 <p><a class="glightbox" href="../intro_to_dynamic_programming.assets/climbing_stairs_dp.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="爬楼梯的动态规划过程" class="animation-figure" src="../intro_to_dynamic_programming.assets/climbing_stairs_dp.png" /></a></p>
 <p align="center"> 图 14-5 &nbsp; 爬楼梯的动态规划过程 </p>
 
-<p>与回溯算法一样，动态规划也使用“状态”概念来表示问题求解的某个特定阶段，每个状态都对应一个子问题以及相应的局部最优解。例如，爬楼梯问题的状态定义为当前所在楼梯阶数 <span class="arithmatex">\(i\)</span> 。</p>
+<p>与回溯算法一样，动态规划也使用“状态”概念来表示问题求解的特定阶段，每个状态都对应一个子问题以及相应的局部最优解。例如，爬楼梯问题的状态定义为当前所在楼梯阶数 <span class="arithmatex">\(i\)</span> 。</p>
 <p>根据以上内容，我们可以总结出动态规划的常用术语。</p>
 <ul>
 <li>将数组 <code>dp</code> 称为「<span class="arithmatex">\(dp\)</span> 表」，<span class="arithmatex">\(dp[i]\)</span> 表示状态 <span class="arithmatex">\(i\)</span> 对应子问题的解。</li>
-<li>将最小子问题对应的状态（即第 <span class="arithmatex">\(1\)</span> 和 <span class="arithmatex">\(2\)</span> 阶楼梯）称为「初始状态」。</li>
+<li>将最小子问题对应的状态（第 <span class="arithmatex">\(1\)</span> 阶和第 <span class="arithmatex">\(2\)</span> 阶楼梯）称为「初始状态」。</li>
 <li>将递推公式 <span class="arithmatex">\(dp[i] = dp[i-1] + dp[i-2]\)</span> 称为「状态转移方程」。</li>
 </ul>
 <h2 id="1414">14.1.4 &nbsp; 空间优化<a class="headerlink" href="#1414" title="Permanent link">&para;</a></h2>
-<p>细心的你可能发现，<strong>由于 <span class="arithmatex">\(dp[i]\)</span> 只与 <span class="arithmatex">\(dp[i-1]\)</span> 和 <span class="arithmatex">\(dp[i-2]\)</span> 有关，因此我们无须使用一个数组 <code>dp</code> 来存储所有子问题的解</strong>，而只需两个变量滚动前进即可。</p>
+<p>细心的读者可能发现了，<strong>由于 <span class="arithmatex">\(dp[i]\)</span> 只与 <span class="arithmatex">\(dp[i-1]\)</span> 和 <span class="arithmatex">\(dp[i-2]\)</span> 有关，因此我们无须使用一个数组 <code>dp</code> 来存储所有子问题的解</strong>，而只需两个变量滚动前进即可。代码如下所示：</p>
 <div class="tabbed-set tabbed-alternate" data-tabs="5:12"><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" /><input id="__tabbed_5_7" name="__tabbed_5" type="radio" /><input id="__tabbed_5_8" name="__tabbed_5" type="radio" /><input id="__tabbed_5_9" name="__tabbed_5" type="radio" /><input id="__tabbed_5_10" name="__tabbed_5" type="radio" /><input id="__tabbed_5_11" name="__tabbed_5" type="radio" /><input id="__tabbed_5_12" name="__tabbed_5" type="radio" /><div class="tabbed-labels"><label for="__tabbed_5_1">Python</label><label for="__tabbed_5_2">C++</label><label for="__tabbed_5_3">Java</label><label for="__tabbed_5_4">C#</label><label for="__tabbed_5_5">Go</label><label for="__tabbed_5_6">Swift</label><label for="__tabbed_5_7">JS</label><label for="__tabbed_5_8">TS</label><label for="__tabbed_5_9">Dart</label><label for="__tabbed_5_10">Rust</label><label for="__tabbed_5_11">C</label><label for="__tabbed_5_12">Zig</label></div>
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@@ -4719,7 +4720,7 @@ dp[i] = dp[i-1] + dp[i-2]
 </div>
 </div>
 </div>
-<p>观察以上代码，由于省去了数组 <code>dp</code> 占用的空间，因此空间复杂度从 <span class="arithmatex">\(O(n)\)</span> 降低至 <span class="arithmatex">\(O(1)\)</span> 。</p>
+<p>观察以上代码，由于省去了数组 <code>dp</code> 占用的空间，因此空间复杂度从 <span class="arithmatex">\(O(n)\)</span> 降至 <span class="arithmatex">\(O(1)\)</span> 。</p>
 <p>在动态规划问题中，当前状态往往仅与前面有限个状态有关，这时我们可以只保留必要的状态，通过“降维”来节省内存空间。<strong>这种空间优化技巧被称为“滚动变量”或“滚动数组”</strong>。</p>
 
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