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第 8 章 &nbsp; 堆
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第 10 章 &nbsp; 搜索
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10.2 &nbsp; 二分查找插入点
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10.5 &nbsp; 重识搜索算法
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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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11.4 &nbsp; 插入排序
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11.5 &nbsp; 快速排序
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11.6 &nbsp; 归并排序
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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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第 12 章 &nbsp; 分治
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第 14 章 &nbsp; 动态规划
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<h1 id="107">10.7 &nbsp; 练习<a class="headerlink" href="#107" title="Permanent link">&para;</a></h1>
<h2 id="1071">10.7.1 &nbsp; 知识巩固<a class="headerlink" href="#1071" title="Permanent link">&para;</a></h2>
<h3 id="1">1. &nbsp; 二分查找怎样缩小搜索区间<a class="headerlink" href="#1" title="Permanent link">&para;</a></h3>
<p>在有序数组 <code>[2, 5, 8, 12, 16, 23, 38]</code> 中查找 16。
使用双闭区间 <code>[i, j]</code>,中点为
<span class="arithmatex">\(m=i+(j-i)/2\)</span>(向下取整)。</p>
<p>请写出每轮的 <code>(i, j, m)</code>、中点元素,以及下一步如何缩小区间,
直到找到目标。</p>
<details class="success">
<summary>参考答案</summary>
<p>每轮查找过程如下:</p>
<table>
<thead>
<tr>
<th>轮次</th>
<th><code>(i, j, m)</code></th>
<th>中点元素</th>
<th>下一步</th>
</tr>
</thead>
<tbody>
<tr>
<td>1</td>
<td><code>(0, 6, 3)</code></td>
<td>12</td>
<td><code>12 &lt; 16</code>,令 <code>i = 4</code></td>
</tr>
<tr>
<td>2</td>
<td><code>(4, 6, 5)</code></td>
<td>23</td>
<td><code>23 &gt; 16</code>,令 <code>j = 4</code></td>
</tr>
<tr>
<td>3</td>
<td><code>(4, 4, 4)</code></td>
<td>16</td>
<td>找到目标,返回索引 4</td>
</tr>
</tbody>
</table>
<p>数组有序,所以中点值小于目标时可以排除中点及其左侧;
中点值大于目标时可以排除中点及其右侧。</p>
</details>
<h3 id="2">2. &nbsp; 重复元素的左右边界<a class="headerlink" href="#2" title="Permanent link">&para;</a></h3>
<p>在数组 <code>[1, 2, 2, 2, 4, 6]</code> 中查找数字 2。
一名同学用二分查找在索引 2 找到目标后立即返回,并说:
“索引 2 就是数字 2 的左边界。”</p>
<!-- numbered-subquestions -->
<ol>
<li>这名同学的说法是否正确?数字 2 的左、右边界分别是多少?请说明理由。</li>
<li>查找左边界时,若中点元素等于目标,接下来应继续搜索哪一侧?</li>
<li>查找右边界时又应继续搜索哪一侧?只需说明方向,无须写出完整的查找过程。</li>
</ol>
<details class="success">
<summary>参考答案</summary>
<ol>
<li>
<p>这名同学的说法不正确。找到一个 2 后立即返回,只能保证找到某一个 2,不能保证找到最左或最右的 2。
本题的左边界为索引 1,右边界为索引 3。</p>
</li>
<li>
<p>查找左边界时,即使中点元素等于 2,也要继续搜索左侧,
例如在双闭区间写法中令 <code>j = m - 1</code>。</p>
</li>
<li>
<p>查找右边界时,中点元素等于 2 后应继续搜索右侧,
例如令 <code>i = m + 1</code>。</p>
</li>
</ol>
</details>
<h3 id="3">3. &nbsp; 不同数据该选哪种搜索方法<a class="headerlink" href="#3" title="Permanent link">&para;</a></h3>
<p>请在“顺序查找、二分查找、哈希表”中,为下面三个场景选择合适的方法,并说明理由:</p>
<!-- numbered-subquestions -->
<ol>
<li>在 <span class="arithmatex">\(10^7\)</span> 个已经有序且不再变动的整数中反复查找,不额外建立其他数据结构;</li>
<li>在频繁插入、删除的数据集中反复判断某个键是否存在,不要求保持有序,也不进行范围查找;</li>
<li>在无序数组中只查找一次某个值。</li>
</ol>
<details class="success">
<summary>参考答案</summary>
<ol>
<li>二分查找:数据有序且静态,<span class="arithmatex">\(O(\log n)\)</span> 且无须额外空间。</li>
<li>哈希表:当哈希函数能把键较均匀地分散到各桶时,插入、删除和按键查找的平均时间复杂度都是 <span class="arithmatex">\(O(1)\)</span>。</li>
<li>
<p>直接从头到尾遍历:只查找一次时,排序或建立哈希表都要先处理整个数组,
并不会减少这一次任务的总工作量。</p>
<p>选择取决于数据是否有序、是否允许建立额外结构、查询次数以及需要执行哪些操作。</p>
</li>
</ol>
</details>
<h2 id="1072">10.7.2 &nbsp; 编程练习<a class="headerlink" href="#1072" title="Permanent link">&para;</a></h2>
<h3 id="1_1">1. &nbsp; 有序数组的二分查找<a class="headerlink" href="#1_1" title="Permanent link">&para;</a></h3>
<p>给定一个按严格递增顺序排列的整数数组 <code>nums</code> 和目标值 <code>target</code>。请使用二分查找找到 <code>target</code>:若存在,返回它的数组索引;若不存在,返回 -1。</p>
<details class="tip">
<summary>解题提示</summary>
<ol>
<li>初始区间是 left = 0、right = n - 1,非空条件是 left &lt;= right</li>
<li>用 mid = left + (right - left) // 2 计算中点</li>
<li>若 <code>nums[mid]</code> 小于 <code>target</code>,就把左边界移到 <code>mid + 1</code>;若 <code>nums[mid]</code> 大于 <code>target</code>,就把右边界移到 <code>mid - 1</code>;相等时立即返回</li>
</ol>
</details>
<p><a class="rounded-button exercise-button" href="https://leetcode.cn/problems/binary-search/" rel="noopener noreferrer" target="_blank">LeetCode</a> <a class="rounded-button exercise-button" href="https://leetcode.cn/problems/binary-search/solutions/1692151/by-jyd-i7xr/" rel="noopener noreferrer" target="_blank">题目解析</a></p>
<h3 id="2_1">2. &nbsp; 有序数组的插入位置<a class="headerlink" href="#2_1" title="Permanent link">&para;</a></h3>
<p>给定一个按严格递增顺序排列的整数数组 <code>nums</code> 和目标值 <code>target</code>。</p>
<p>若 <code>target</code> 已在数组中,返回它的索引;否则,返回把 <code>target</code> 插入数组后仍能保持严格递增顺序的位置。答案可能是 0,也可能等于数组长度。请使用二分查找。</p>
<details class="tip">
<summary>解题提示</summary>
<ol>
<li>答案可能是 0,也可能是数组长度 n</li>
<li>使用双闭区间时,若 <code>nums[mid]</code> 大于等于 <code>target</code>,就令 <code>right = mid - 1</code>,继续检查更靠左的位置;否则令 <code>left = mid + 1</code></li>
<li>循环结束时 left 就是插入位置</li>
</ol>
</details>
<p><a class="rounded-button exercise-button" href="https://leetcode.cn/problems/search-insert-position/" rel="noopener noreferrer" target="_blank">LeetCode</a></p>
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