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第 1 章 &nbsp; 初识算法
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第 6 章 &nbsp; 哈希表
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第 8 章 &nbsp; 堆
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第 9 章 &nbsp; 图
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10.2 &nbsp; 二分查找插入点
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10.3 &nbsp; 二分查找边界
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10.4 &nbsp; 哈希优化策略
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第 11 章 &nbsp; 排序
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11.2 &nbsp; 选择排序
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11.3 &nbsp; 冒泡排序
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11.4 &nbsp; 插入排序
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11.5 &nbsp; 快速排序
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11.7 &nbsp; 堆排序
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11.8 &nbsp; 桶排序
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11.9 &nbsp; 计数排序
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11.10 &nbsp; 基数排序
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第 12 章 &nbsp; 分治
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12.1 &nbsp; 分治算法
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12.4 &nbsp; 汉诺塔问题
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第 13 章 &nbsp; 回溯
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第 14 章 &nbsp; 动态规划
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第 15 章 &nbsp; 贪心
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第 16 章 &nbsp; 附录
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<h1 id="33">3.3 &nbsp; 数字编码 *<a class="headerlink" href="#33" title="Permanent link">&para;</a></h1>
<div class="admonition note">
<p class="admonition-title">Note</p>
<p>在本书中，标题带有 * 符号的是选读章节。如果你时间有限或感到理解困难，可以先跳过，等学完必读章节后再单独攻克。</p>
</div>
<h2 id="331">3.3.1 &nbsp; 原码、反码和补码<a class="headerlink" href="#331" title="Permanent link">&para;</a></h2>
<p>在上一节的表格中我们发现，所有整数类型能够表示的负数都比正数多一个，例如 <code>byte</code> 的取值范围是 <span class="arithmatex">\([-128, 127]\)</span> 。这个现象比较反直觉，它的内在原因涉及原码、反码、补码的相关知识。</p>
<p>首先需要指出，<strong>数字是以“补码”的形式存储在计算机中的</strong>。在分析这样做的原因之前，首先给出三者的定义。</p>
<ul>
<li><strong>原码</strong>：我们将数字的二进制表示的最高位视为符号位，其中 <span class="arithmatex">\(0\)</span> 表示正数，<span class="arithmatex">\(1\)</span> 表示负数，其余位表示数字的值。</li>
<li><strong>反码</strong>：正数的反码与其原码相同，负数的反码是对其原码除符号位外的所有位取反。</li>
<li><strong>补码</strong>：正数的补码与其原码相同，负数的补码是在其反码的基础上加 <span class="arithmatex">\(1\)</span> 。</li>
</ul>
<p>图 3-4 展示了原码、反码和补码之间的转换方法。</p>
<p><a class="glightbox" href="../number_encoding.assets/1s_2s_complement.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="原码、反码与补码之间的相互转换" class="animation-figure" src="../number_encoding.assets/1s_2s_complement.png" /></a></p>
<p align="center"> 图 3-4 &nbsp; 原码、反码与补码之间的相互转换 </p>
<p>「原码 sign-magnitude」虽然最直观，但存在一些局限性。一方面，<strong>负数的原码不能直接用于运算</strong>。例如在原码下计算 <span class="arithmatex">\(1 + (-2)\)</span> ，得到的结果是 <span class="arithmatex">\(-3\)</span> ，这显然是不对的。</p>
<div class="arithmatex">\[
\begin{aligned}
&amp; 1 + (-2) \newline
&amp; \rightarrow 0000 \; 0001 + 1000 \; 0010 \newline
&amp; = 1000 \; 0011 \newline
&amp; \rightarrow -3
\end{aligned}
\]</div>
<p>为了解决此问题，计算机引入了「反码 1's complement」。如果我们先将原码转换为反码，并在反码下计算 <span class="arithmatex">\(1 + (-2)\)</span> ，最后将结果从反码转换回原码，则可得到正确结果 <span class="arithmatex">\(-1\)</span> 。</p>
<div class="arithmatex">\[
\begin{aligned}
&amp; 1 + (-2) \newline
&amp; \rightarrow 0000 \; 0001 \; \text{(原码)} + 1000 \; 0010 \; \text{(原码)} \newline
&amp; = 0000 \; 0001 \; \text{(反码)} + 1111 \; 1101 \; \text{(反码)} \newline
&amp; = 1111 \; 1110 \; \text{(反码)} \newline
&amp; = 1000 \; 0001 \; \text{(原码)} \newline
&amp; \rightarrow -1
\end{aligned}
\]</div>
<p>另一方面，<strong>数字零的原码有 <span class="arithmatex">\(+0\)</span> 和 <span class="arithmatex">\(-0\)</span> 两种表示方式</strong>。这意味着数字零对应两个不同的二进制编码，这可能会带来歧义。比如在条件判断中，如果没有区分正零和负零，则可能会导致判断结果出错。而如果我们想处理正零和负零歧义，则需要引入额外的判断操作，这可能会降低计算机的运算效率。</p>
<div class="arithmatex">\[
\begin{aligned}
+0 &amp; \rightarrow 0000 \; 0000 \newline
-0 &amp; \rightarrow 1000 \; 0000
\end{aligned}
\]</div>
<p>与原码一样，反码也存在正负零歧义问题，因此计算机进一步引入了「补码 2's complement」。我们先来观察一下负零的原码、反码、补码的转换过程：</p>
<div class="arithmatex">\[
\begin{aligned}
-0 \rightarrow \; &amp; 1000 \; 0000 \; \text{(原码)} \newline
= \; &amp; 1111 \; 1111 \; \text{(反码)} \newline
= 1 \; &amp; 0000 \; 0000 \; \text{(补码)} \newline
\end{aligned}
\]</div>
<p>在负零的反码基础上加 <span class="arithmatex">\(1\)</span> 会产生进位，但 <code>byte</code> 类型的长度只有 8 位，因此溢出到第 9 位的 <span class="arithmatex">\(1\)</span> 会被舍弃。也就是说，<strong>负零的补码为 <span class="arithmatex">\(0000 \; 0000\)</span> ，与正零的补码相同</strong>。这意味着在补码表示中只存在一个零，正负零歧义从而得到解决。</p>
<p>还剩最后一个疑惑：<code>byte</code> 类型的取值范围是 <span class="arithmatex">\([-128, 127]\)</span> ，多出来的一个负数 <span class="arithmatex">\(-128\)</span> 是如何得到的呢？我们注意到，区间 <span class="arithmatex">\([-127, +127]\)</span> 内的所有整数都有对应的原码、反码和补码，并且原码和补码之间可以互相转换。</p>
<p>然而，<strong>补码 <span class="arithmatex">\(1000 \; 0000\)</span> 是一个例外，它并没有对应的原码</strong>。根据转换方法，我们得到该补码的原码为 <span class="arithmatex">\(0000 \; 0000\)</span> 。这显然是矛盾的，因为该原码表示数字 <span class="arithmatex">\(0\)</span> ，它的补码应该是自身。计算机规定这个特殊的补码 <span class="arithmatex">\(1000 \; 0000\)</span> 代表 <span class="arithmatex">\(-128\)</span> 。实际上，<span class="arithmatex">\((-1) + (-127)\)</span> 在补码下的计算结果就是 <span class="arithmatex">\(-128\)</span> 。</p>
<div class="arithmatex">\[
\begin{aligned}
&amp; (-127) + (-1) \newline
&amp; \rightarrow 1111 \; 1111 \; \text{(原码)} + 1000 \; 0001 \; \text{(原码)} \newline
&amp; = 1000 \; 0000 \; \text{(反码)} + 1111 \; 1110 \; \text{(反码)} \newline
&amp; = 1000 \; 0001 \; \text{(补码)} + 1111 \; 1111 \; \text{(补码)} \newline
&amp; = 1000 \; 0000 \; \text{(补码)} \newline
&amp; \rightarrow -128
\end{aligned}
\]</div>
<p>你可能已经发现了，上述所有计算都是加法运算。这暗示着一个重要事实：<strong>计算机内部的硬件电路主要是基于加法运算设计的</strong>。这是因为加法运算相对于其他运算（比如乘法、除法和减法）来说，硬件实现起来更简单，更容易进行并行化处理，运算速度更快。</p>
<p>请注意，这并不意味着计算机只能做加法。<strong>通过将加法与一些基本逻辑运算结合，计算机能够实现各种其他的数学运算</strong>。例如，计算减法 <span class="arithmatex">\(a - b\)</span> 可以转换为计算加法 <span class="arithmatex">\(a + (-b)\)</span> ；计算乘法和除法可以转换为计算多次加法或减法。</p>
<p>现在我们可以总结出计算机使用补码的原因：基于补码表示，计算机可以用同样的电路和操作来处理正数和负数的加法，不需要设计特殊的硬件电路来处理减法，并且无须特别处理正负零的歧义问题。这大大简化了硬件设计，提高了运算效率。</p>
<p>补码的设计非常精妙，因篇幅关系我们就先介绍到这里，建议有兴趣的读者进一步深入了解。</p>
<h2 id="332">3.3.2 &nbsp; 浮点数编码<a class="headerlink" href="#332" title="Permanent link">&para;</a></h2>
<p>细心的你可能会发现：<code>int</code> 和 <code>float</code> 长度相同，都是 4 字节 ，但为什么 <code>float</code> 的取值范围远大于 <code>int</code> ？这非常反直觉，因为按理说 <code>float</code> 需要表示小数，取值范围应该变小才对。</p>
<p>实际上，<strong>这是因为浮点数 <code>float</code> 采用了不同的表示方式</strong>。记一个 32 位长度的二进制数为：</p>
<div class="arithmatex">\[
b_{31} b_{30} b_{29} \ldots b_2 b_1 b_0
\]</div>
<p>根据 IEEE 754 标准，32-bit 长度的 <code>float</code> 由以下三个部分构成。</p>
<ul>
<li>符号位 <span class="arithmatex">\(\mathrm{S}\)</span> ：占 1 位 ，对应 <span class="arithmatex">\(b_{31}\)</span> 。</li>
<li>指数位 <span class="arithmatex">\(\mathrm{E}\)</span> ：占 8 位 ，对应 <span class="arithmatex">\(b_{30} b_{29} \ldots b_{23}\)</span> 。</li>
<li>分数位 <span class="arithmatex">\(\mathrm{N}\)</span> ：占 23 位 ，对应 <span class="arithmatex">\(b_{22} b_{21} \ldots b_0\)</span> 。</li>
</ul>
<p>二进制数 <code>float</code> 对应值的计算方法为：</p>
<div class="arithmatex">\[
\text {val} = (-1)^{b_{31}} \times 2^{\left(b_{30} b_{29} \ldots b_{23}\right)_2-127} \times\left(1 . b_{22} b_{21} \ldots b_0\right)_2
\]</div>
<p>转化到十进制下的计算公式为：</p>
<div class="arithmatex">\[
\text {val}=(-1)^{\mathrm{S}} \times 2^{\mathrm{E} -127} \times (1 + \mathrm{N})
\]</div>
<p>其中各项的取值范围为：</p>
<div class="arithmatex">\[
\begin{aligned}
\mathrm{S} \in &amp; \{ 0, 1\}, \quad \mathrm{E} \in \{ 1, 2, \dots, 254 \} \newline
(1 + \mathrm{N}) = &amp; (1 + \sum_{i=1}^{23} b_{23-i} 2^{-i}) \subset [1, 2 - 2^{-23}]
\end{aligned}
\]</div>
<p><a class="glightbox" href="../number_encoding.assets/ieee_754_float.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="IEEE 754 标准下的 float 的计算示例" class="animation-figure" src="../number_encoding.assets/ieee_754_float.png" /></a></p>
<p align="center"> 图 3-5 &nbsp; IEEE 754 标准下的 float 的计算示例 </p>
<p>观察图 3-5 ，给定一个示例数据 <span class="arithmatex">\(\mathrm{S} = 0\)</span> ， <span class="arithmatex">\(\mathrm{E} = 124\)</span> ，<span class="arithmatex">\(\mathrm{N} = 2^{-2} + 2^{-3} = 0.375\)</span> ，则有：</p>
<div class="arithmatex">\[
\text { val } = (-1)^0 \times 2^{124 - 127} \times (1 + 0.375) = 0.171875
\]</div>
<p>现在我们可以回答最初的问题：<strong><code>float</code> 的表示方式包含指数位，导致其取值范围远大于 <code>int</code></strong> 。根据以上计算，<code>float</code> 可表示的最大正数为 <span class="arithmatex">\(2^{254 - 127} \times (2 - 2^{-23}) \approx 3.4 \times 10^{38}\)</span> ，切换符号位便可得到最小负数。</p>
<p><strong>尽管浮点数 <code>float</code> 扩展了取值范围，但其副作用是牺牲了精度</strong>。整数类型 <code>int</code> 将全部 32 位用于表示数字，数字是均匀分布的；而由于指数位的存在，浮点数 <code>float</code> 的数值越大，相邻两个数字之间的差值就会趋向越大。</p>
<p>如表 3-2 所示，指数位 <span class="arithmatex">\(E = 0\)</span> 和 <span class="arithmatex">\(E = 255\)</span> 具有特殊含义，<strong>用于表示零、无穷大、<span class="arithmatex">\(\mathrm{NaN}\)</span> 等</strong>。</p>
<p align="center"> 表 3-2 &nbsp; 指数位含义 </p>
<div class="center-table">
<table>
<thead>
<tr>
<th>指数位 E</th>
<th>分数位 <span class="arithmatex">\(\mathrm{N} = 0\)</span></th>
<th>分数位 <span class="arithmatex">\(\mathrm{N} \ne 0\)</span></th>
<th>计算公式</th>
</tr>
</thead>
<tbody>
<tr>
<td><span class="arithmatex">\(0\)</span></td>
<td><span class="arithmatex">\(\pm 0\)</span></td>
<td>次正规数</td>
<td><span class="arithmatex">\((-1)^{\mathrm{S}} \times 2^{-126} \times (0.\mathrm{N})\)</span></td>
</tr>
<tr>
<td><span class="arithmatex">\(1, 2, \dots, 254\)</span></td>
<td>正规数</td>
<td>正规数</td>
<td><span class="arithmatex">\((-1)^{\mathrm{S}} \times 2^{(\mathrm{E} -127)} \times (1.\mathrm{N})\)</span></td>
</tr>
<tr>
<td><span class="arithmatex">\(255\)</span></td>
<td><span class="arithmatex">\(\pm \infty\)</span></td>
<td><span class="arithmatex">\(\mathrm{NaN}\)</span></td>
<td></td>
</tr>
</tbody>
</table>
</div>
<p>值得说明的是，次正规数显著提升了浮点数的精度。最小正正规数为 <span class="arithmatex">\(2^{-126}\)</span> ，最小正次正规数为 <span class="arithmatex">\(2^{-126} \times 2^{-23}\)</span> 。</p>
<p>双精度 <code>double</code> 也采用类似于 <code>float</code> 的表示方法，在此不做赘述。</p>
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