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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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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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2.6   练习
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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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3.6   练习
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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   内存与缓存 *
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4.5   小结
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4.6   练习
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4.6   练习
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4.6.1   知识巩固
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1.   数组和链表怎样找到元素
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2.   数组和链表怎样插入元素
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3.   列表容量是怎样增长的
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4.6.2   编程练习
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1.   数组表示的大整数加一
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2.   反转单向链表
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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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5.5   练习
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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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6.5   练习
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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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7.7   练习
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第 8 章   堆
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第 8 章   堆
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<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_heap/heap/">
<span class="md-ellipsis">
8.1   堆
</span>
</a>
</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_heap/build_heap/">
<span class="md-ellipsis">
8.2   建堆操作
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</a>
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<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_heap/top_k/">
<span class="md-ellipsis">
8.3   Top-k 问题
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<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_heap/summary/">
<span class="md-ellipsis">
8.4   小结
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</a>
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<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_heap/exercises/">
<span class="md-ellipsis">
8.5   练习
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第 9 章   图
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第 9 章   图
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<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_graph/graph/">
<span class="md-ellipsis">
9.1   图
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</a>
</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_graph/graph_operations/">
<span class="md-ellipsis">
9.2   图基础操作
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</a>
</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_graph/graph_traversal/">
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9.3   图的遍历
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<a class="md-nav__link" href="../../chapter_graph/summary/">
<span class="md-ellipsis">
9.4   小结
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</a>
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<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_graph/exercises/">
<span class="md-ellipsis">
9.5   练习
</span>
</a>
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<a class="md-nav__link" href="../../chapter_searching/">
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第 10 章   搜索
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第 10 章   搜索
</label>
<ul class="md-nav__list" data-md-scrollfix="">
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_searching/binary_search/">
<span class="md-ellipsis">
10.1   二分查找
</span>
</a>
</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_searching/binary_search_insertion/">
<span class="md-ellipsis">
10.2   二分查找插入点
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</a>
</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_searching/binary_search_edge/">
<span class="md-ellipsis">
10.3   二分查找边界
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</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_searching/replace_linear_by_hashing/">
<span class="md-ellipsis">
10.4   哈希优化策略
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</a>
</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_searching/searching_algorithm_revisited/">
<span class="md-ellipsis">
10.5   重识搜索算法
</span>
</a>
</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_searching/summary/">
<span class="md-ellipsis">
10.6   小结
</span>
</a>
</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_searching/exercises/">
<span class="md-ellipsis">
10.7   练习
</span>
</a>
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</ul>
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第 11 章   排序
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第 11 章   排序
</label>
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<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_sorting/sorting_algorithm/">
<span class="md-ellipsis">
11.1   排序算法
</span>
</a>
</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_sorting/selection_sort/">
<span class="md-ellipsis">
11.2   选择排序
</span>
</a>
</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_sorting/bubble_sort/">
<span class="md-ellipsis">
11.3   冒泡排序
</span>
</a>
</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_sorting/insertion_sort/">
<span class="md-ellipsis">
11.4   插入排序
</span>
</a>
</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_sorting/quick_sort/">
<span class="md-ellipsis">
11.5   快速排序
</span>
</a>
</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_sorting/merge_sort/">
<span class="md-ellipsis">
11.6   归并排序
</span>
</a>
</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_sorting/heap_sort/">
<span class="md-ellipsis">
11.7   堆排序
</span>
</a>
</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_sorting/bucket_sort/">
<span class="md-ellipsis">
11.8   桶排序
</span>
</a>
</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_sorting/counting_sort/">
<span class="md-ellipsis">
11.9   计数排序
</span>
</a>
</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_sorting/radix_sort/">
<span class="md-ellipsis">
11.10   基数排序
</span>
</a>
</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_sorting/summary/">
<span class="md-ellipsis">
11.11   小结
</span>
</a>
</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_sorting/exercises/">
<span class="md-ellipsis">
11.12   练习
</span>
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第 12 章   分治
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第 12 章   分治
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<a class="md-nav__link" href="../../chapter_divide_and_conquer/divide_and_conquer/">
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12.1   分治算法
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</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_divide_and_conquer/binary_search_recur/">
<span class="md-ellipsis">
12.2   分治搜索策略
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</a>
</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_divide_and_conquer/build_binary_tree_problem/">
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12.3   构建树问题
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</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_divide_and_conquer/hanota_problem/">
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12.4   汉诺塔问题
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</a>
</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_divide_and_conquer/summary/">
<span class="md-ellipsis">
12.5   小结
</span>
</a>
</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_divide_and_conquer/exercises/">
<span class="md-ellipsis">
12.6   练习
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第 13 章   回溯
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第 13 章   回溯
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<a class="md-nav__link" href="../../chapter_backtracking/backtracking_algorithm/">
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13.1   回溯算法
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<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_backtracking/permutations_problem/">
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13.2   全排列问题
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<a class="md-nav__link" href="../../chapter_backtracking/subset_sum_problem/">
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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 class="md-nav__link" href="../../chapter_backtracking/exercises/">
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13.6   练习
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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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<a class="md-nav__link" href="../../chapter_dynamic_programming/intro_to_dynamic_programming/">
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14.1   初探动态规划
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</a>
</li>
<li class="md-nav__item">
<a class="md-nav__link" href="../../chapter_dynamic_programming/dp_problem_features/">
<span class="md-ellipsis">
14.2   DP 问题特性
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<a class="md-nav__link" href="../../chapter_dynamic_programming/dp_solution_pipeline/">
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14.3   DP 解题思路
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</li>
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<a class="md-nav__link" href="../../chapter_dynamic_programming/knapsack_problem/">
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14.4   0-1 背包问题
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<a class="md-nav__link" href="../../chapter_dynamic_programming/unbounded_knapsack_problem/">
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14.5   完全背包问题
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<a class="md-nav__link" href="../../chapter_dynamic_programming/edit_distance_problem/">
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14.6   编辑距离问题
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<a class="md-nav__link" href="../../chapter_dynamic_programming/summary/">
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14.7   小结
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14.8   练习
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<h1 id="46">4.6   练习<a class="headerlink" href="#46" title="Permanent link"></a></h1>
<h2 id="461">4.6.1   知识巩固<a class="headerlink" href="#461" title="Permanent link"></a></h2>
<h3 id="1">1.   数组和链表怎样找到元素<a class="headerlink" href="#1" title="Permanent link"></a></h3>
<p>数组和单向链表中都按顺序保存了 <code>[A, B, C, D, E]</code>。现在要读取第 4 个元素 <code>D</code>:</p>
<!-- numbered-subquestions -->
<ol>
<li>数组可以直接使用哪个索引?</li>
<li>单向链表从头节点 <code>A</code> 开始,要沿着 <code>next</code> 依次经过哪些节点?</li>
<li>当要读取的元素位置越来越靠后时,两种结构所需的步骤会怎样变化?哪一种更适合反复按位置读取?为什么?</li>
</ol>
<details class="success">
<summary>参考答案</summary>
<ol>
<li>
<p>若索引从 0 开始,第 4 个元素的索引是 3,数组可直接读取 <code>arr[3]</code>。</p>
</li>
<li>
<p>单向链表必须从头开始,访问路径为 <code>A → B → C → D</code>,需要沿 <code>next</code> 前进 3 次。</p>
</li>
<li>
<p>数组可以根据首地址和索引直接定位元素,按位置访问的时间复杂度为 <span class="arithmatex">\(O(1)\)</span>。
单向链表访问第 <span class="arithmatex">\(k\)</span> 个节点时,必须从头节点开始,沿着 <code>next</code> 前进 <span class="arithmatex">\(k-1\)</span> 次,
最坏需要 <span class="arithmatex">\(O(n)\)</span> 时间。</p>
<p>这里只比较按位置读取,不表示链表在所有操作上都更慢。</p>
</li>
</ol>
</details>
<h3 id="2">2.   数组和链表怎样插入元素<a class="headerlink" href="#2" title="Permanent link"></a></h3>
<p>数组和单向链表中都保存了 <code>A、B、C、D</code>。现在要在 <code>B</code> 后面插入 <code>X</code>:</p>
<ul>
<li>数组容量为 5,当前状态是 <code>[A, B, C, D, _]</code>;</li>
<li>链表为 <code>A → B → C → D</code>,并且已经拿到了指向节点 <code>B</code> 的引用。</li>
</ul>
<!-- numbered-subquestions -->
<ol>
<li>数组需要移动哪些元素?写出插入后的数组。</li>
<li>链表应按什么顺序修改 <code>X.next</code> 和 <code>B.next</code>?写出插入后的链表。</li>
<li>为什么比较插入效率时,要特别说明“已经拿到了节点 B 的引用”?</li>
</ol>
<details class="success">
<summary>参考答案</summary>
<ol>
<li>
<p>数组要先把 <code>D</code> 向右移动一格,再把 <code>C</code> 向右移动一格,最后把 <code>X</code> 放入索引 2,
得到 <code>[A, B, X, C, D]</code>。</p>
</li>
<li>
<p><code>B.next</code> 原本指向 <code>C</code>。应先令 <code>X.next = B.next</code>,让 <code>X</code> 指向 <code>C</code>;
再令 <code>B.next = X</code>。结果为 <code>A → B → X → C → D</code>。
如果先覆盖 <code>B.next</code> 又没有保存原来的连接,就可能找不到 <code>C</code>。</p>
</li>
<li>
<p>已知 <code>B</code> 的位置后,链表插入只需修改两个连接,可以在 <span class="arithmatex">\(O(1)\)</span> 时间完成。
如果还要从头查找 <code>B</code>,查找过程本身可能需要 <span class="arithmatex">\(O(n)\)</span> 时间。</p>
</li>
</ol>
</details>
<h3 id="3">3.   列表容量是怎样增长的<a class="headerlink" href="#3" title="Permanent link"></a></h3>
<p>一个基于数组实现的列表当前内容为 <code>[A, B, C]</code>,长度 <code>size = 3</code>,容量 <code>capacity = 4</code>。
规定容量不足时,新数组的容量扩大为原来的 2 倍。</p>
<!-- numbered-subquestions -->
<ol>
<li>追加 <code>D</code> 后,列表的长度和容量分别是多少?是否需要扩容?</li>
<li>接着追加 <code>E</code> 时,容量会变为多少?需要复制几个原有元素?</li>
<li>底层数组的长度不可变,为什么列表的容量看起来却可以增长?</li>
</ol>
<details class="success">
<summary>参考答案</summary>
<ol>
<li>
<p><code>D</code> 可以放入最后一个空位。此时内容为 <code>[A, B, C, D]</code>,
<code>size = 4</code>、<code>capacity = 4</code>,不需要扩容。</p>
</li>
<li>
<p>再追加 <code>E</code> 时已经没有空位,需要创建容量为 8 的新数组,
将 <code>A、B、C、D</code> 共 4 个原有元素复制过去,再加入 <code>E</code>。
此时 <code>size = 5</code>、<code>capacity = 8</code>。</p>
</li>
<li>
<p>原来的数组本身没有变长。列表创建了一个更大的新数组,复制原有元素,
再改用新数组作为底层存储,因此对使用者来说容量增长了。</p>
</li>
</ol>
</details>
<h2 id="462">4.6.2   编程练习<a class="headerlink" href="#462" title="Permanent link"></a></h2>
<h3 id="1_1">1.   数组表示的大整数加一<a class="headerlink" href="#1_1" title="Permanent link"></a></h3>
<p>数组 <code>digits</code> 从左到右保存一个非负整数的各位数字,例如 <code>[3, 0, 8]</code> 表示 308。
数字 0 用 <code>[0]</code> 表示;其他输入的第一位均不为 0。</p>
<p>请模拟一次十进制竖式加法,将这个整数增加 1,并把结果仍按相同的数组形式返回。
可以直接修改 <code>digits</code>;若最前面产生新的进位,可以返回一个更长的数组。</p>
<details class="tip">
<summary>解题提示</summary>
<ol>
<li>像做竖式加法一样,从数组最后一位开始</li>
<li>当前位小于 9 时加一后即可立即返回</li>
<li>当前位等于 9 时把它改成 0;若所有位都是 9,需要在最前面补 1</li>
</ol>
</details>
<p><a class="rounded-button exercise-button" href="https://leetcode.cn/problems/plus-one/" rel="noopener noreferrer" target="_blank">LeetCode</a></p>
<h3 id="2_1">2.   反转单向链表<a class="headerlink" href="#2_1" title="Permanent link"></a></h3>
<p>给定一个单向链表的头节点 <code>head</code>。每个节点包含一个值和指向下一节点的 <code>next</code>。</p>
<p>请使用迭代方法反转所有节点之间的连接,并返回反转后的头节点。
要求不创建新的链表节点。</p>
<details class="tip">
<summary>解题提示</summary>
<ol>
<li>先在纸上画出三个相连的节点和 prev、cur 两个指针</li>
<li>改写 cur.next 前,必须先用 nxt 保存原来的下一个节点</li>
<li>反转 cur.next 后,令 prev = cur,再令 cur = nxt,继续处理原链表中的下一个节点</li>
</ol>
</details>
<p><a class="rounded-button exercise-button" href="https://leetcode.cn/problems/reverse-linked-list/" rel="noopener noreferrer" target="_blank">LeetCode</a> <a class="rounded-button exercise-button" href="https://leetcode.cn/problems/reverse-linked-list/solutions/2361282/206-fan-zhuan-lian-biao-shuang-zhi-zhen-r1jel/" rel="noopener noreferrer" target="_blank">题目解析</a></p>
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