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3. 数据结构
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3.1. 数据结构分类
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3.2. 基本数据类型
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3.3. 数字编码 *
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< a href = "./" class = "md-nav__link md-nav__link--active" >
3.3. 数字编码 *
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目录
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3.3.1. 原码、反码和补码
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< a href = "#332" class = "md-nav__link" >
3.3.2. 浮点数编码
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3.4. 字符编码 *
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3.5. 小结
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4. 数组与链表
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4.1. 数组
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4.2. 链表
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4.3. 列表
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4.4. 小结
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5.1. 栈
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5.2. 队列
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5.3. 双向队列
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5.4. 小结
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6. 散列表
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6.1. 哈希表（ ）
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6.2. 哈希冲突（ ）
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6.3. 哈希算法（ ）
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6.4. 小结
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7. 树
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7.1. 二叉树
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7.2. 二叉树遍历
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7.3. 二叉树数组表示
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7.4. 二叉搜索树
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7.5. AVL 树 *
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7.6. 小结
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8. 堆
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8.1. 堆
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< a href = "../../chapter_heap/build_heap/" class = "md-nav__link" >
8.2. 建堆操作
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8.3. Top-K 问题（ ）
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8.4. 小结
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9.1. 图
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9.2. 图基础操作
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9.3. 图的遍历
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9.4. 小结
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10. 搜索
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10.1. 二分查找
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10.2. 二分查找边界
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10.3. 哈希优化策略
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10.4. 重识搜索算法
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10.5. 小结
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11. 排序
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11.1. 排序算法
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11.2. 选择排序
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11.3. 冒泡排序
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11.4. 插入排序
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11.5. 快速排序
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11.6. 归并排序
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11.7. 堆排序
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11.8. 桶排序
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11.9. 计数排序
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11.10. 基数排序
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11.11. 小结
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3.3.1. 原码、反码和补码
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3.3.2. 浮点数编码
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< h1 id = "33" > 3.3. 数字编码 *< a class = "headerlink" href = "#33" title = "Permanent link" > ¶ < / a > < / h1 >
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< p class = "admonition-title" > Note< / p >
< p > 在本书中，标题带有的 * 符号的是选读章节。如果你时间有限或感到理解困难，建议先跳过，等学完必读章节后再单独攻克。< / p >
< / div >
< h2 id = "331" > 3.3.1. 原码、反码和补码< a class = "headerlink" href = "#331" title = "Permanent link" > ¶ < / a > < / h2 >
< p > 从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，它的内在原因涉及到原码、反码、补码的相关知识。在展开分析之前，我们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 > < img alt = "原码、反码与补码之间的相互转换" src = "../number_encoding.assets/1s_2s_complement.png" / > < / p >
< p align = "center" > Fig. 原码、反码与补码之间的相互转换 < / p >
< p > 显然，「原码」最为直观，< strong > 然而数字却是以「补码」的形式存储在计算机中的< / strong > 。这是因为原码存在一些局限性。< / p >
< p > 一方面，< strong > 负数的原码不能直接用于运算< / strong > 。例如，我们在原码下计算 < 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" > \[
\begin{aligned}
& 1 + (-2) \newline
& = 0000 \space 0001 + 1000 \space 0010 \newline
& = 1000 \space 0011 \newline
& = -3
\end{aligned}
\]< / div >
< p > 为了解决此问题，计算机引入了「反码」。例如，我们先将原码转换为反码，并在反码下计算 < span class = "arithmatex" > \(1 + (-2)\)< / span > ，并将结果从反码转化回原码，则可得到正确结果 < span class = "arithmatex" > \(-1\)< / span > 。< / p >
< div class = "arithmatex" > \[
\begin{aligned}
& 1 + (-2) \newline
& = 0000 \space 0001 \space \text{(原码)} + 1000 \space 0010 \space \text{(原码)} \newline
& = 0000 \space 0001 \space \text{(反码)} + 1111 \space 1101 \space \text{(反码)} \newline
& = 1111 \space 1110 \space \text{(反码)} \newline
& = 1000 \space 0001 \space \text{(原码)} \newline
& = -1
\end{aligned}
\]< / 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着两个不同的二进制编码，而这可能会带来歧义问题。例如，在条件判断中，如果没有区分正零和负零，可能会导致错误的判断结果。如果我们想要处理正零和负零歧义，则需要引入额外的判断操作，其可能会降低计算机的运算效率。< / p >
< div class = "arithmatex" > \[
\begin{aligned}
+0 & = 0000 \space 0000 \newline
-0 & = 1000 \space 0000
\end{aligned}
\]< / div >
< p > 与原码一样，反码也存在正负零歧义问题。为此，计算机进一步引入了「补码」。那么，补码有什么作用呢？我们先来分析一下负零的补码的计算过程：< / p >
< div class = "arithmatex" > \[
\begin{aligned}
-0 = \space & 1000 \space 0000 \space \text{(原码)} \newline
= \space & 1111 \space 1111 \space \text{(反码)} \newline
= 1 \space & 0000 \space 0000 \space \text{(补码)} \newline
\end{aligned}
\]< / div >
< p > 在负零的反码基础上加 < span class = "arithmatex" > \(1\)< / span > 会产生进位，而由于 byte 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。< strong > 从而得到负零的补码为 < span class = "arithmatex" > \(0000 \space 0000\)< / 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，从而解决了正负零歧义问题。< / p >
< p > 还剩余最后一个疑惑： < 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是可以互相转换的。< / p >
< p > 然而，< strong > 补码 < span class = "arithmatex" > \(1000 \space 0000\)< / 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" > \(0000 \space 0000\)< / 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" > \(1000 \space 0000\)< / 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" > \[
\begin{aligned}
& (-127) + (-1) \newline
& = 1111 \space 1111 \space \text{(原码)} + 1000 \space 0001 \space \text{(原码)} \newline
& = 1000 \space 0000 \space \text{(反码)} + 1111 \space 1110 \space \text{(反码)} \newline
& = 1000 \space 0001 \space \text{(补码)} + 1111 \space 1111 \space \text{(补码)} \newline
& = 1000 \space 0000 \space \text{(补码)} \newline
& = -128
\end{aligned}
\]< / div >
< p > 你可能已经发现，上述的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，从而提高运算速度。< / p >
< p > 然而，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，并且无需特别处理正负零的歧义问题。这大大简化了硬件设计，并提高了运算效率。< / p >
< p > 补码的设计非常精妙，由于篇幅关系我们先介绍到这里。建议有兴趣的读者进一步深度了解。< / p >
< h2 id = "332" > 3.3.2. 浮点数编码< a class = "headerlink" href = "#332" title = "Permanent link" > ¶ < / 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 bytes， < 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，这是因为浮点数 < code > float< / code > 采用了不同的表示方式。根据 IEEE 754 标准， < code > float< / code > 由以下部分构成：< / p >
< ul >
< li > 符号位 < span class = "arithmatex" > \(\mathrm{S}\)< / span > ：占 1 bit ； < / li >
< li > 指数位 < span class = "arithmatex" > \(\mathrm{E}\)< / span > ：占 8 bits ； < / li >
< li > 分数位 < span class = "arithmatex" > \(\mathrm{N}\)< / span > ：占 24 bits ，其中 23 位显式存储；< / li >
< / ul >
< p > 设 32-bit 二进制数的第 < span class = "arithmatex" > \(i\)< / span > 位为 < span class = "arithmatex" > \(b_i\)< / span > ，则 < code > float< / code > 值的计算方法定义为：< / 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" > \[
\begin{aligned}
\mathrm{S} \in & \{ 0, 1\} , \quad \mathrm{E} \in \{ 1, 2, \dots, 254 \} \newline
(1 + \mathrm{N}) = & (1 + \sum_{i=1}^{23} b_{23-i} 2^{-i}) \subset [1, 2 - 2^{-23}]
\end{aligned}
\]< / div >
< p > < img alt = "IEEE 754 标准下的 float 表示方式" src = "../number_encoding.assets/ieee_754_float.png" / > < / p >
< p align = "center" > Fig. IEEE 754 标准下的 float 表示方式 < / p >
< p > 以上图为例，< 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 > ，易得< / 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 > 进一步地，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 >
< 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 > 特别地，次正规数显著提升了浮点数的精度，这是因为：< / p >
< ul >
< li > 最小正正规数为 < span class = "arithmatex" > \(2^{-126} \approx 1.18 \times 10^{-38}\)< / span > ； < / li >
< li > 最小正次正规数为 < span class = "arithmatex" > \(2^{-126} \times 2^{-23} \approx 1.4 \times 10^{-45}\)< / span > ； < / li >
< / ul >
< p > 双精度 < code > double< / code > 也采用类似 < code > float< / code > 的表示方法，此处不再详述。< / p >
< h2 id = "__comments" > 评论< / h2 >
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