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chapter_divide_and_conquer/binary_search_recur/index.html
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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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第 12 章 &nbsp; 分治
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12.1 &nbsp; 分治算法
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12.3 &nbsp; 构建树问题
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15.4 &nbsp; 最大切分乘积问题
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<li>循环第 <code>1.</code> 和 <code>2.</code> 步，直至找到 <code>target</code> 或区间为空时返回。</li>
</ol>
<p>图 12-4 展示了在数组中二分查找元素 <span class="arithmatex">\(6\)</span> 的分治过程。</p>
<p><img alt="二分查找的分治过程" src="../binary_search_recur.assets/binary_search_recur.png" /></p>
<p><a class="glightbox" href="../binary_search_recur.assets/binary_search_recur.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="二分查找的分治过程" src="../binary_search_recur.assets/binary_search_recur.png" /></a></p>
<p align="center"> 图 12-4 &nbsp; 二分查找的分治过程 </p>
<p>在实现代码中，我们声明一个递归函数 <code>dfs()</code> 来求解问题 <span class="arithmatex">\(f(i, j)\)</span> 。</p>
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12.3 &nbsp; 构建树问题
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<p class="admonition-title">Question</p>
<p>给定一个二叉树的前序遍历 <code>preorder</code> 和中序遍历 <code>inorder</code> ，请从中构建二叉树，返回二叉树的根节点。</p>
</div>
<p><img alt="构建二叉树的示例数据" src="../build_binary_tree_problem.assets/build_tree_example.png" /></p>
<p><a class="glightbox" href="../build_binary_tree_problem.assets/build_tree_example.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="构建二叉树的示例数据" src="../build_binary_tree_problem.assets/build_tree_example.png" /></a></p>
<p align="center"> 图 12-5 &nbsp; 构建二叉树的示例数据 </p>
<h3 id="1">1. &nbsp; 判断是否为分治问题<a class="headerlink" href="#1" title="Permanent link">&para;</a></h3>
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<li>查找根节点 3 在 <code>inorder</code> 中的索引，利用该索引可将 <code>inorder</code> 划分为 <code>[ 9 | 3 ｜ 1 2 7 ]</code> 。</li>
<li>根据 <code>inorder</code> 划分结果，易得左子树和右子树的节点数量分别为 1 和 3 ，从而可将 <code>preorder</code> 划分为 <code>[ 3 | 9 | 2 1 7 ]</code> 。</li>
</ol>
<p><img alt="在前序和中序遍历中划分子树" src="../build_binary_tree_problem.assets/build_tree_preorder_inorder_division.png" /></p>
<p><a class="glightbox" href="../build_binary_tree_problem.assets/build_tree_preorder_inorder_division.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="在前序和中序遍历中划分子树" src="../build_binary_tree_problem.assets/build_tree_preorder_inorder_division.png" /></a></p>
<p align="center"> 图 12-6 &nbsp; 在前序和中序遍历中划分子树 </p>
<h3 id="3">3. &nbsp; 基于变量描述子树区间<a class="headerlink" href="#3" title="Permanent link">&para;</a></h3>
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</table>
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<p>请注意，右子树根节点索引中的 <span class="arithmatex">\((m-l)\)</span> 的含义是“左子树的节点数量”，建议配合图 12-7 理解。</p>
<p><img alt="根节点和左右子树的索引区间表示" src="../build_binary_tree_problem.assets/build_tree_division_pointers.png" /></p>
<p><a class="glightbox" href="../build_binary_tree_problem.assets/build_tree_division_pointers.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="根节点和左右子树的索引区间表示" src="../build_binary_tree_problem.assets/build_tree_division_pointers.png" /></a></p>
<p align="center"> 图 12-7 &nbsp; 根节点和左右子树的索引区间表示 </p>
<h3 id="4">4. &nbsp; 代码实现<a class="headerlink" href="#4" title="Permanent link">&para;</a></h3>
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<div class="tabbed-set tabbed-alternate" data-tabs="2:9"><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" /><div class="tabbed-labels"><label for="__tabbed_2_1">&lt;1&gt;</label><label for="__tabbed_2_2">&lt;2&gt;</label><label for="__tabbed_2_3">&lt;3&gt;</label><label for="__tabbed_2_4">&lt;4&gt;</label><label for="__tabbed_2_5">&lt;5&gt;</label><label for="__tabbed_2_6">&lt;6&gt;</label><label for="__tabbed_2_7">&lt;7&gt;</label><label for="__tabbed_2_8">&lt;8&gt;</label><label for="__tabbed_2_9">&lt;9&gt;</label></div>
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<p><img alt="构建二叉树的递归过程" src="../build_binary_tree_problem.assets/built_tree_step1.png" /></p>
<p><a class="glightbox" href="../build_binary_tree_problem.assets/built_tree_step1.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="构建二叉树的递归过程" src="../build_binary_tree_problem.assets/built_tree_step1.png" /></a></p>
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<p><img alt="built_tree_step2" src="../build_binary_tree_problem.assets/built_tree_step2.png" /></p>
<p><a class="glightbox" href="../build_binary_tree_problem.assets/built_tree_step2.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="built_tree_step2" src="../build_binary_tree_problem.assets/built_tree_step2.png" /></a></p>
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<p><img alt="built_tree_step3" src="../build_binary_tree_problem.assets/built_tree_step3.png" /></p>
<p><a class="glightbox" href="../build_binary_tree_problem.assets/built_tree_step3.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="built_tree_step3" src="../build_binary_tree_problem.assets/built_tree_step3.png" /></a></p>
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<p><img alt="built_tree_step4" src="../build_binary_tree_problem.assets/built_tree_step4.png" /></p>
<p><a class="glightbox" href="../build_binary_tree_problem.assets/built_tree_step4.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="built_tree_step4" src="../build_binary_tree_problem.assets/built_tree_step4.png" /></a></p>
</div>
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<p><img alt="built_tree_step5" src="../build_binary_tree_problem.assets/built_tree_step5.png" /></p>
<p><a class="glightbox" href="../build_binary_tree_problem.assets/built_tree_step5.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="built_tree_step5" src="../build_binary_tree_problem.assets/built_tree_step5.png" /></a></p>
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<p><img alt="built_tree_step6" src="../build_binary_tree_problem.assets/built_tree_step6.png" /></p>
<p><a class="glightbox" href="../build_binary_tree_problem.assets/built_tree_step6.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="built_tree_step6" src="../build_binary_tree_problem.assets/built_tree_step6.png" /></a></p>
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<p><img alt="built_tree_step7" src="../build_binary_tree_problem.assets/built_tree_step7.png" /></p>
<p><a class="glightbox" href="../build_binary_tree_problem.assets/built_tree_step7.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="built_tree_step7" src="../build_binary_tree_problem.assets/built_tree_step7.png" /></a></p>
</div>
<div class="tabbed-block">
<p><img alt="built_tree_step8" src="../build_binary_tree_problem.assets/built_tree_step8.png" /></p>
<p><a class="glightbox" href="../build_binary_tree_problem.assets/built_tree_step8.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="built_tree_step8" src="../build_binary_tree_problem.assets/built_tree_step8.png" /></a></p>
</div>
<div class="tabbed-block">
<p><img alt="built_tree_step9" src="../build_binary_tree_problem.assets/built_tree_step9.png" /></p>
<p><a class="glightbox" href="../build_binary_tree_problem.assets/built_tree_step9.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="built_tree_step9" src="../build_binary_tree_problem.assets/built_tree_step9.png" /></a></p>
</div>
</div>
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<p align="center"> 图 12-8 &nbsp; 构建二叉树的递归过程 </p>
<p>每个递归函数内的前序遍历 <code>preorder</code> 和中序遍历 <code>inorder</code> 的划分结果如图 12-9 所示。</p>
<p><img alt="每个递归函数中的划分结果" src="../build_binary_tree_problem.assets/built_tree_overall.png" /></p>
<p><a class="glightbox" href="../build_binary_tree_problem.assets/built_tree_overall.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="每个递归函数中的划分结果" src="../build_binary_tree_problem.assets/built_tree_overall.png" /></a></p>
<p align="center"> 图 12-9 &nbsp; 每个递归函数中的划分结果 </p>
<p>设树的节点数量为 <span class="arithmatex">\(n\)</span> ，初始化每一个节点（执行一个递归函数 <code>dfs()</code> ）使用 <span class="arithmatex">\(O(1)\)</span> 时间。<strong>因此总体时间复杂度为 <span class="arithmatex">\(O(n)\)</span></strong> 。</p>
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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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12.3 &nbsp; 构建树问题
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<li><strong>分</strong>：递归地将原数组（原问题）划分为两个子数组（子问题），直到子数组只剩一个元素（最小子问题）。</li>
<li><strong>治</strong>：从底至顶地将有序的子数组（子问题的解）进行合并，从而得到有序的原数组（原问题的解）。</li>
</ol>
<p><img alt="归并排序的分治策略" src="../divide_and_conquer.assets/divide_and_conquer_merge_sort.png" /></p>
<p><a class="glightbox" href="../divide_and_conquer.assets/divide_and_conquer_merge_sort.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="归并排序的分治策略" src="../divide_and_conquer.assets/divide_and_conquer_merge_sort.png" /></a></p>
<p align="center"> 图 12-1 &nbsp; 归并排序的分治策略 </p>
<h2 id="1211">12.1.1 &nbsp; 如何判断分治问题<a class="headerlink" href="#1211" title="Permanent link">&para;</a></h2>
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<div class="arithmatex">\[
O(n + (\frac{n}{2})^2 \times 2 + n) = O(\frac{n^2}{2} + 2n)
\]</div>
<p><img alt="划分数组前后的冒泡排序" src="../divide_and_conquer.assets/divide_and_conquer_bubble_sort.png" /></p>
<p><a class="glightbox" href="../divide_and_conquer.assets/divide_and_conquer_bubble_sort.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="划分数组前后的冒泡排序" src="../divide_and_conquer.assets/divide_and_conquer_bubble_sort.png" /></a></p>
<p align="center"> 图 12-2 &nbsp; 划分数组前后的冒泡排序 </p>
<p>接下来，我们计算以下不等式，其左边和右边分别为划分前和划分后的操作总数：</p>
@@ -3588,7 +3415,7 @@ n(n - 4) &amp; &gt; 0
<p>我们知道，分治生成的子问题是相互独立的，<strong>因此通常可以并行解决</strong>。也就是说，分治不仅可以降低算法的时间复杂度，<strong>还有利于操作系统的并行优化</strong>。</p>
<p>并行优化在多核或多处理器的环境中尤其有效，因为系统可以同时处理多个子问题，更加充分地利用计算资源，从而显著减少总体的运行时间。</p>
<p>比如在图 12-3 所示的“桶排序”中，我们将海量的数据平均分配到各个桶中，则可所有桶的排序任务分散到各个计算单元，完成后再进行结果合并。</p>
<p><img alt="桶排序的并行计算" src="../divide_and_conquer.assets/divide_and_conquer_parallel_computing.png" /></p>
<p><a class="glightbox" href="../divide_and_conquer.assets/divide_and_conquer_parallel_computing.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="桶排序的并行计算" src="../divide_and_conquer.assets/divide_and_conquer_parallel_computing.png" /></a></p>
<p align="center"> 图 12-3 &nbsp; 桶排序的并行计算 </p>
<h2 id="1213">12.1.3 &nbsp; 分治常见应用<a class="headerlink" href="#1213" title="Permanent link">&para;</a></h2>
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html.glightbox-open { overflow: initial; height: 100%; }
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10.2 &nbsp; 二分查找插入点
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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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12.2 &nbsp; 分治搜索策略
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12.3 &nbsp; 构建树问题
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12.4 &nbsp; 汉诺塔问题
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第 14 章 &nbsp; 动态规划
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14.2 &nbsp; DP 问题特性
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<li>小圆盘必须时刻位于大圆盘之上。</li>
</ol>
</div>
<p><img alt="汉诺塔问题示例" src="../hanota_problem.assets/hanota_example.png" /></p>
<p><a class="glightbox" href="../hanota_problem.assets/hanota_example.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="汉诺塔问题示例" src="../hanota_problem.assets/hanota_example.png" /></a></p>
<p align="center"> 图 12-10 &nbsp; 汉诺塔问题示例 </p>
<p><strong>我们将规模为 <span class="arithmatex">\(i\)</span> 的汉诺塔问题记做 <span class="arithmatex">\(f(i)\)</span></strong> 。例如 <span class="arithmatex">\(f(3)\)</span> 代表将 <span class="arithmatex">\(3\)</span> 个圆盘从 <code>A</code> 移动至 <code>C</code> 的汉诺塔问题。</p>
@@ -3515,10 +3342,10 @@
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<div class="tabbed-content">
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<p><img alt="规模为 1 问题的解" src="../hanota_problem.assets/hanota_f1_step1.png" /></p>
<p><a class="glightbox" href="../hanota_problem.assets/hanota_f1_step1.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="规模为 1 问题的解" src="../hanota_problem.assets/hanota_f1_step1.png" /></a></p>
</div>
<div class="tabbed-block">
<p><img alt="hanota_f1_step2" src="../hanota_problem.assets/hanota_f1_step2.png" /></p>
<p><a class="glightbox" href="../hanota_problem.assets/hanota_f1_step2.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="hanota_f1_step2" src="../hanota_problem.assets/hanota_f1_step2.png" /></a></p>
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@@ -3533,16 +3360,16 @@
<div class="tabbed-set tabbed-alternate" data-tabs="2:4"><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" /><div class="tabbed-labels"><label for="__tabbed_2_1">&lt;1&gt;</label><label for="__tabbed_2_2">&lt;2&gt;</label><label for="__tabbed_2_3">&lt;3&gt;</label><label for="__tabbed_2_4">&lt;4&gt;</label></div>
<div class="tabbed-content">
<div class="tabbed-block">
<p><img alt="规模为 2 问题的解" src="../hanota_problem.assets/hanota_f2_step1.png" /></p>
<p><a class="glightbox" href="../hanota_problem.assets/hanota_f2_step1.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="规模为 2 问题的解" src="../hanota_problem.assets/hanota_f2_step1.png" /></a></p>
</div>
<div class="tabbed-block">
<p><img alt="hanota_f2_step2" src="../hanota_problem.assets/hanota_f2_step2.png" /></p>
<p><a class="glightbox" href="../hanota_problem.assets/hanota_f2_step2.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="hanota_f2_step2" src="../hanota_problem.assets/hanota_f2_step2.png" /></a></p>
</div>
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<p><img alt="hanota_f2_step3" src="../hanota_problem.assets/hanota_f2_step3.png" /></p>
<p><a class="glightbox" href="../hanota_problem.assets/hanota_f2_step3.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="hanota_f2_step3" src="../hanota_problem.assets/hanota_f2_step3.png" /></a></p>
</div>
<div class="tabbed-block">
<p><img alt="hanota_f2_step4" src="../hanota_problem.assets/hanota_f2_step4.png" /></p>
<p><a class="glightbox" href="../hanota_problem.assets/hanota_f2_step4.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="hanota_f2_step4" src="../hanota_problem.assets/hanota_f2_step4.png" /></a></p>
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@@ -3560,16 +3387,16 @@
<div class="tabbed-set tabbed-alternate" data-tabs="3:4"><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" /><div class="tabbed-labels"><label for="__tabbed_3_1">&lt;1&gt;</label><label for="__tabbed_3_2">&lt;2&gt;</label><label for="__tabbed_3_3">&lt;3&gt;</label><label for="__tabbed_3_4">&lt;4&gt;</label></div>
<div class="tabbed-content">
<div class="tabbed-block">
<p><img alt="规模为 3 问题的解" src="../hanota_problem.assets/hanota_f3_step1.png" /></p>
<p><a class="glightbox" href="../hanota_problem.assets/hanota_f3_step1.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="规模为 3 问题的解" src="../hanota_problem.assets/hanota_f3_step1.png" /></a></p>
</div>
<div class="tabbed-block">
<p><img alt="hanota_f3_step2" src="../hanota_problem.assets/hanota_f3_step2.png" /></p>
<p><a class="glightbox" href="../hanota_problem.assets/hanota_f3_step2.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="hanota_f3_step2" src="../hanota_problem.assets/hanota_f3_step2.png" /></a></p>
</div>
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<p><img alt="hanota_f3_step3" src="../hanota_problem.assets/hanota_f3_step3.png" /></p>
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<li>将 <span class="arithmatex">\(n-1\)</span> 个圆盘借助 <code>A</code> 从 <code>B</code> 移至 <code>C</code> 。</li>
</ol>
<p>对于这两个子问题 <span class="arithmatex">\(f(n-1)\)</span> ，<strong>可以通过相同的方式进行递归划分</strong>，直至达到最小子问题 <span class="arithmatex">\(f(1)\)</span> 。而 <span class="arithmatex">\(f(1)\)</span> 的解是已知的，只需一次移动操作即可。</p>
<p><img alt="汉诺塔问题的分治策略" src="../hanota_problem.assets/hanota_divide_and_conquer.png" /></p>
<p><a class="glightbox" href="../hanota_problem.assets/hanota_divide_and_conquer.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="汉诺塔问题的分治策略" src="../hanota_problem.assets/hanota_divide_and_conquer.png" /></a></p>
<p align="center"> 图 12-14 &nbsp; 汉诺塔问题的分治策略 </p>
<h3 id="3">3. &nbsp; 代码实现<a class="headerlink" href="#3" title="Permanent link">&para;</a></h3>
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<p>如图 12-15 所示，汉诺塔问题形成一个高度为 <span class="arithmatex">\(n\)</span> 的递归树，每个节点代表一个子问题、对应一个开启的 <code>dfs()</code> 函数，<strong>因此时间复杂度为 <span class="arithmatex">\(O(2^n)\)</span> ，空间复杂度为 <span class="arithmatex">\(O(n)\)</span></strong> 。</p>
<p><img alt="汉诺塔问题的递归树" src="../hanota_problem.assets/hanota_recursive_tree.png" /></p>
<p><a class="glightbox" href="../hanota_problem.assets/hanota_recursive_tree.png" data-type="image" data-width="100%" data-height="auto" data-desc-position="bottom"><img alt="汉诺塔问题的递归树" src="../hanota_problem.assets/hanota_recursive_tree.png" /></a></p>
<p align="center"> 图 12-15 &nbsp; 汉诺塔问题的递归树 </p>
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15.1 &nbsp; 贪心算法
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15.4 &nbsp; 最大切分乘积问题
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<h1 id="12">第 12 章 &nbsp; 分治<a class="headerlink" href="#12" title="Permanent link">&para;</a></h1>
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@@ -3592,10 +3427,15 @@ aria-label="页脚"
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第 14 章 &nbsp; 动态规划
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