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跳转至
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Hello 算法
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7.6 小结
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第 0 章 前言
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第 0 章 前言
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0.1 关于本书
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0.2 如何使用本书
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0.3 小结
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第 1 章 初识算法
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第 1 章 初识算法
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1.1 算法无处不在
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1.2 算法是什么
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1.3 小结
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< a href = "../../chapter_computational_complexity/" class = "md-nav__link " >
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第 2 章 复杂度分析
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第 2 章 复杂度分析
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2.1 算法效率评估
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2.2 迭代与递归
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2.3 时间复杂度
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2.4 空间复杂度
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2.5 小结
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< a href = "../../chapter_data_structure/" class = "md-nav__link " >
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第 3 章 数据结构
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第 3 章 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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3.4 字符编码 *
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3.5 小结
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< a href = "../../chapter_array_and_linkedlist/" class = "md-nav__link " >
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第 4 章 数组与链表
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第 4 章 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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< a href = "../../chapter_stack_and_queue/" class = "md-nav__link " >
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第 5 章 栈与队列
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第 5 章 栈与队列
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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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< a href = "../../chapter_hashing/" class = "md-nav__link " >
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第 6 章 哈希表
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第 6 章 哈希表
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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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< a href = "../" class = "md-nav__link " >
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第 7 章 树
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第 7 章 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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7.6 小结
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目录
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1. 重点回顾
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2. Q & A
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< a href = "../../chapter_heap/" class = "md-nav__link " >
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第 8 章 堆
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第 8 章 堆
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8.1 堆
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< a href = "../../chapter_heap/build_heap/" class = "md-nav__link" >
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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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< a href = "../../chapter_graph/" class = "md-nav__link " >
< svg xmlns = "http://www.w3.org/2000/svg" viewBox = "0 0 24 24" > < path d = "m12 5.37-.44-.06L6 14.9c.24.21.4.48.47.78h11.06c.07-.3.23-.57.47-.78l-5.56-9.59-.44.06M6.6 16.53l4.28 2.53c.29-.27.69-.43 1.12-.43.43 0 .83.16 1.12.43l4.28-2.53H6.6M12 22a1.68 1.68 0 0 1-1.68-1.68l.09-.56-4.3-2.55c-.31.36-.76.58-1.27.58a1.68 1.68 0 0 1-1.68-1.68c0-.79.53-1.45 1.26-1.64V9.36c-.83-.11-1.47-.82-1.47-1.68A1.68 1.68 0 0 1 4.63 6c.55 0 1.03.26 1.34.66l4.41-2.53-.06-.45c0-.93.75-1.68 1.68-1.68.93 0 1.68.75 1.68 1.68l-.06.45 4.41 2.53c.31-.4.79-.66 1.34-.66a1.68 1.68 0 0 1 1.68 1.68c0 .86-.64 1.57-1.47 1.68v5.11c.73.19 1.26.85 1.26 1.64a1.68 1.68 0 0 1-1.68 1.68c-.51 0-.96-.22-1.27-.58l-4.3 2.55.09.56A1.68 1.68 0 0 1 12 22M10.8 4.86 6.3 7.44l.02.24c0 .71-.44 1.32-1.06 1.57l.03 5.25 5.51-9.64m2.4 0 5.51 9.64.03-5.25c-.62-.25-1.06-.86-1.06-1.57l.02-.24-4.5-2.58Z" / > < / svg >
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第 9 章 图
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第 9 章 图
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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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< a href = "../../chapter_searching/" class = "md-nav__link " >
< svg xmlns = "http://www.w3.org/2000/svg" viewBox = "0 0 24 24" > < path d = "m19.31 18.9 3.08 3.1L21 23.39l-3.12-3.07c-.69.43-1.51.68-2.38.68-2.5 0-4.5-2-4.5-4.5s2-4.5 4.5-4.5 4.5 2 4.5 4.5c0 .88-.25 1.71-.69 2.4m-3.81.1a2.5 2.5 0 0 0 0-5 2.5 2.5 0 0 0 0 5M21 4v2H3V4h18M3 16v-2h6v2H3m0-5V9h18v2h-2.03c-1.01-.63-2.2-1-3.47-1s-2.46.37-3.47 1H3Z" / > < / svg >
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第 10 章 搜索
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第 10 章 搜索
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10.1 二分查找
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10.2 二分查找插入点
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10.3 二分查找边界
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10.4 哈希优化策略
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10.5 重识搜索算法
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10.6 小结
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< a href = "../../chapter_sorting/" class = "md-nav__link " >
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第 11 章 排序
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第 11 章 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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< a href = "../../chapter_divide_and_conquer/" class = "md-nav__link " >
< svg xmlns = "http://www.w3.org/2000/svg" viewBox = "0 0 24 24" > < path d = "M17 7v2h5V7h-5M2 9v6h5V9H2m10 0v2H9v2h3v2l3-3-3-3m5 2v2h5v-2h-5m0 4v2h5v-2h-5Z" / > < / svg >
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第 12 章 分治
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第 12 章 分治
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12.1 分治算法
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12.2 分治搜索策略
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12.3 构建树问题
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12.4 汉诺塔问题
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12.5 小结
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< a href = "../../chapter_backtracking/" class = "md-nav__link " >
< svg xmlns = "http://www.w3.org/2000/svg" viewBox = "0 0 24 24" > < path d = "M18 15a3 3 0 0 1 3 3 3 3 0 0 1-3 3 2.99 2.99 0 0 1-2.83-2H14v-2h1.17c.41-1.17 1.52-2 2.83-2m0 2a1 1 0 0 0-1 1 1 1 0 0 0 1 1 1 1 0 0 0 1-1 1 1 0 0 0-1-1m0-9a1.43 1.43 0 0 0 1.43-1.43 1.43 1.43 0 1 0-2.86 0A1.43 1.43 0 0 0 18 8m0-5.43a4 4 0 0 1 4 4C22 9.56 18 14 18 14s-4-4.44-4-7.43a4 4 0 0 1 4-4M8.83 17H10v2H8.83A2.99 2.99 0 0 1 6 21a3 3 0 0 1-3-3c0-1.31.83-2.42 2-2.83V14h2v1.17c.85.3 1.53.98 1.83 1.83M6 17a1 1 0 0 0-1 1 1 1 0 0 0 1 1 1 1 0 0 0 1-1 1 1 0 0 0-1-1M6 3a3 3 0 0 1 3 3c0 1.31-.83 2.42-2 2.83V10H5V8.83A2.99 2.99 0 0 1 3 6a3 3 0 0 1 3-3m0 2a1 1 0 0 0-1 1 1 1 0 0 0 1 1 1 1 0 0 0 1-1 1 1 0 0 0-1-1m5 14v-2h2v2h-2m-4-6H5v-2h2v2Z" / > < / svg >
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第 13 章 回溯
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第 13 章 回溯
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13.1 回溯算法
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13.2 全排列问题
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13.3 子集和问题
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13.4 N 皇后问题
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13.5 小结
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< a href = "../../chapter_dynamic_programming/" class = "md-nav__link " >
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第 14 章 动态规划
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第 14 章 动态规划
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14.1 初探动态规划
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14.2 DP 问题特性
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14.3 DP 解题思路
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14.4 0-1 背包问题
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14.5 完全背包问题
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14.6 编辑距离问题
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14.7 小结
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< a href = "../../chapter_greedy/" class = "md-nav__link " >
< svg xmlns = "http://www.w3.org/2000/svg" viewBox = "0 0 24 24" > < path d = "M13 3c3.88 0 7 3.14 7 7 0 2.8-1.63 5.19-4 6.31V21H9v-3H8c-1.11 0-2-.89-2-2v-3H4.5c-.42 0-.66-.5-.42-.81L6 9.66A7.003 7.003 0 0 1 13 3m0-2C8.41 1 4.61 4.42 4.06 8.9L2.5 11h-.03l-.02.03c-.55.76-.62 1.76-.19 2.59.36.69 1 1.17 1.74 1.32V16c0 1.85 1.28 3.42 3 3.87V23h11v-5.5c2.5-1.67 4-4.44 4-7.5 0-4.97-4.04-9-9-9m4 7.83c0 1.54-1.36 2.77-3.42 4.64L13 14l-.58-.53C10.36 11.6 9 10.37 9 8.83c0-1.2.96-2.19 2.16-2.2h.04c.69 0 1.35.31 1.8.83.45-.52 1.11-.83 1.8-.83 1.2-.01 2.2.96 2.2 2.16v.04Z" / > < / svg >
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第 15 章 贪心
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< ul >
< li > 二叉树是一种非线性数据结构，体现“一分为二”的分治逻辑。每个二叉树节点包含一个值以及两个指针，分别指向其左子节点和右子节点。< / li >
< li > 对于二叉树中的某个节点，其左（右）子节点及其以下形成的树被称为该节点的左（右）子树。< / li >
< li > 二叉树的相关术语包括根节点、叶节点、层、度、边、高度和深度等。< / li >
< li > 二叉树的初始化、节点插入和节点删除操作与链表操作方法类似。< / li >
< li > 常见的二叉树类型有完美二叉树、完全二叉树、满二叉树和平衡二叉树。完美二叉树是最理想的状态，而链表是退化后的最差状态。< / li >
< li > 二叉树可以用数组表示，方法是将节点值和空位按层序遍历顺序排列，并根据父节点与子节点之间的索引映射关系来实现指针。< / li >
< li > 二叉树的层序遍历是一种广度优先搜索方法，它体现了“一圈一圈向外”的分层遍历方式，通常通过队列来实现。< / li >
< li > 前序、中序、后序遍历皆属于深度优先搜索，它们体现了“走到尽头，再回头继续”的回溯遍历方式，通常使用递归来实现。< / li >
< li > 二叉搜索树是一种高效的元素查找数据结构，其查找、插入和删除操作的时间复杂度均为 < span class = "arithmatex" > \(O(\log n)\)< / span > 。当二叉搜索树退化为链表时，各项时间复杂度会劣化至 < span class = "arithmatex" > \(O(n)\)< / span > 。< / li >
< li > AVL 树，也称为平衡二叉搜索树，它通过旋转操作，确保在不断插入和删除节点后，树仍然保持平衡。< / li >
< li > AVL 树的旋转操作包括右旋、左旋、先右旋再左旋、先左旋再右旋。在插入或删除节点后， < / li >
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< p class = "admonition-title" > 对于只有一个节点的二叉树，树的高度和根节点的深度都是 < span class = "arithmatex" > \(0\)< / span > 吗？< / p >
< p > 是的，因为高度和深度通常定义为“走过边的数量”。< / p >
< / div >
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< p class = "admonition-title" > 二叉树中的插入与删除一般都是由一套操作配合完成的，这里的“一套操作”指什么呢？可以理解为资源的子节点的资源释放吗？< / p >
< p > 拿二叉搜索树来举例，删除节点操作要分为三种情况处理，其中每种情况都需要进行多个步骤的节点操作。< / p >
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< p class = "admonition-title" > 为什么 DFS 遍历二叉树有前、中、后三种顺序，分别有什么用呢？< / p >
< p > DFS 的前、中、后序遍历和访问数组的顺序类似，是遍历二叉树的基本方法，利用这三种遍历方法，我们可以得到一个特定顺序的遍历结果。例如在二叉搜索树中，由于结点大小满足 < code > 左子结点值 < 根结点值 < 右子结点值< / code > ，因此我们只要按照 < code > 左-> 根-> 右< / code > 的优先级遍历树，就可以获得有序的节点序列。< / p >
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< p class = "admonition-title" > 右旋操作是处理失衡节点 < code > node< / code > 、< code > child< / code > 、< code > grand_child< / code > 之间的关系，那 < code > node< / code > 的父节点和 < code > node< / code > 原来的连接不需要维护吗？右旋操作后岂不是断掉了？< / p >
< p > 我们需要从递归的视角来看这个问题。右旋操作 < code > right_rotate(root)< / code > 传入的是子树的根节点，最终 < code > return child< / code > 返回旋转之后的子树的根节点。子树的根节点和其父节点的连接是在该函数返回后完成的，不属于右旋操作的维护范围。< / p >
< / div >
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< p class = "admonition-title" > 在 C++ 中，函数被划分到 < code > private< / code > 和 < code > public< / code > 中，这方面有什么考量吗？为什么要将 < code > height()< / code > 函数和 < code > updateHeight()< / code > 函数分别放在 < code > public< / code > 和 < code > private< / code > 中呢？< / p >
< p > 主要看方法的使用范围，如果方法只在类内部使用，那么就设计为 < code > private< / code > 。例如，用户单独调用 < code > updateHeight()< / code > 是没有意义的，它只是插入、删除操作中的一步。而 < code > height()< / code > 是访问结点高度，类似于 < code > vector.size()< / code > ，因此设置成 < code > public< / code > 以便使用。< / p >
< / div >
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< p class = "admonition-title" > 请问如何从一组输入数据构建一个二叉搜索树？根节点的选择是不是很重要？< / p >
< p > 是的，构建树的方法已在二叉搜索树代码中的 < code > build_tree()< / code > 方法中给出。至于根节点的选择，我们通常会将输入数据排序，然后用中点元素作为根节点，再递归地构建左右子树。这样做可以最大程度保证树的平衡性。< / p >
< / div >
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< p class = "admonition-title" > 在 Java 中，字符串对比是否一定要用 < code > equals()< / code > 方法？< / p >
< p > 在 Java 中，对于基本数据类型，< code > ==< / code > 用于对比两个变量的值是否相等。对于引用类型，两种符号的工作原理是不同的。< / p >
< ul >
< li > < code > ==< / code > ：用来比较两个变量是否指向同一个对象，即它们在内存中的位置是否相同。< / li >
< li > < code > equals()< / code > ：用来对比两个对象的值是否相等。< / li >
< / ul >
< p > 因此如果要对比值，我们通常会用 < code > equals()< / code > 。然而，通过 < code > String a = "hi"; String b = "hi";< / code > 初始化的字符串都存储在字符串常量池中，它们指向同一个对象，因此也可以用 < code > a == b< / code > 来比较两个字符串的内容。< / p >
< / div >
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< p class = "admonition-title" > 广度优先遍历到最底层之前，队列中的节点数量是 < span class = "arithmatex" > \(2^h\)< / span > 吗？< / p >
< p > 是的，例如高度 < span class = "arithmatex" > \(h = 2\)< / span > 的满二叉树，其节点总数 < span class = "arithmatex" > \(n = 7\)< / span > ，则底层节点数量 < span class = "arithmatex" > \(4 = 2^h = (n + 1) / 2\)< / span > 。< / p >
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