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第 8 章 &nbsp; 堆積
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第 9 章 &nbsp; 圖
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第 10 章 &nbsp; 搜尋
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第 11 章 &nbsp; 排序
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第 12 章 &nbsp; 分治
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12.1 &nbsp; 分治演算法
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12.3 &nbsp; 構建樹問題
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第 13 章 &nbsp; 回溯
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14.1 &nbsp; 初探動態規劃
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1. &nbsp; 比較兩個字串的字元組成
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<!-- 此檔案由 utils/exercises/publish_exercises.py 根據 exercises.yaml 自動生成，請勿直接修改。 -->
<h1 id="65">6.5 &nbsp; 練習<a class="headerlink" href="#65" title="Permanent link">&para;</a></h1>
<h2 id="651">6.5.1 &nbsp; 知識鞏固<a class="headerlink" href="#651" title="Permanent link">&para;</a></h2>
<h3 id="1">1. &nbsp; 發生雜湊衝突後怎樣查詢<a class="headerlink" href="#1" title="Permanent link">&para;</a></h3>
<p>一個雜湊表有 5 個桶，雜湊函式為 <span class="arithmatex">\(h(x)=x \bmod 5\)</span>，衝突時把元素依次放進該桶的串列中。
依次插入 <code>[1, 6, 11, 7]</code>：</p>
<!-- numbered-subquestions -->
<ol>
<li>寫出 0～4 號桶中的內容；</li>
<li>查詢 6 時會先進入哪個桶，並依次檢查哪些元素？</li>
<li>根據第 1 問寫出的桶內容，後插入的元素是否覆蓋了先插入的元素？結合這種衝突處理方式說明理由。</li>
</ol>
<details class="success">
<summary>參考答案</summary>
<ol>
<li>
<p>因為 <span class="arithmatex">\(1\bmod5=6\bmod5=11\bmod5=1\)</span>，而 <span class="arithmatex">\(7\bmod5=2\)</span>，各桶為：</p>
<div class="highlight"><pre><span></span><code><a id="__codelineno-0-1" name="__codelineno-0-1" href="#__codelineno-0-1"></a>0: []
<a id="__codelineno-0-2" name="__codelineno-0-2" href="#__codelineno-0-2"></a>1: [1, 6, 11]
<a id="__codelineno-0-3" name="__codelineno-0-3" href="#__codelineno-0-3"></a>2: [7]
<a id="__codelineno-0-4" name="__codelineno-0-4" href="#__codelineno-0-4"></a>3: []
<a id="__codelineno-0-5" name="__codelineno-0-5" href="#__codelineno-0-5"></a>4: []
</code></pre></div>
</li>
<li>
<p>查詢 6 時先進入 1 號桶，再依次比較 1、6，第二次比較時找到目標。</p>
</li>
<li>
<p>雜湊值相同只表示它們落入同一個桶，並不表示這些元素相等。鏈式位址把衝突元素都儲存在桶內，
查詢時再逐個比較，因此 1、6、11 不會互相覆蓋。</p>
</li>
</ol>
</details>
<h3 id="2">2. &nbsp; 雜湊表擴容後元素去哪兒<a class="headerlink" href="#2" title="Permanent link">&para;</a></h3>
<p>一個使用鏈式位址的雜湊表原有 5 個桶，雜湊函式為 <span class="arithmatex">\(h(x)=x\bmod5\)</span>。
鍵 <code>[1, 6, 11]</code> 都在 1 號桶中。</p>
<p>現在把雜湊表擴容為 7 個桶，雜湊函式相應變為 <span class="arithmatex">\(h(x)=x\bmod7\)</span>：</p>
<!-- numbered-subquestions -->
<ol>
<li>分別計算 1、6、11 的新桶號。</li>
<li>擴容後哪些桶中有元素？</li>
<li>擴容時能否把原來 1 號桶中的串列原樣複製到新的 1 號桶？結合第 1、2 問的結果說明理由。</li>
</ol>
<details class="success">
<summary>參考答案</summary>
<ol>
<li>
<p>新桶號分別為：</p>
<ul>
<li><span class="arithmatex">\(1\bmod7=1\)</span>；</li>
<li><span class="arithmatex">\(6\bmod7=6\)</span>；</li>
<li><span class="arithmatex">\(11\bmod7=4\)</span>。</li>
</ul>
</li>
<li>
<p>1 號桶儲存 1，4 號桶儲存 11，6 號桶儲存 6。三個鍵不再擠在同一個桶中。</p>
</li>
<li>
<p>不能原樣複製。桶號由“鍵對桶數取餘”得到。桶數從 5 變成 7 後，同一個鍵的新桶號可能改變，
因此必須重新計算每個鍵的位置。若把舊的 1 號桶原樣複製過去，之後按新公式查詢 6 和 11 時，
會分別前往 6 號桶和 4 號桶，從而找不到它們。</p>
</li>
</ol>
</details>
<h3 id="3-6-11">3. &nbsp; 刪除 6 後還能找到 11 嗎<a class="headerlink" href="#3-6-11" title="Permanent link">&para;</a></h3>
<p>一個雜湊表有 5 個位置，索引為 <code>0～4</code>，雜湊函式為 <span class="arithmatex">\(h(x)=x\bmod5\)</span>。
發生衝突時，從雜湊函式算出的索引開始，向右尋找第一個空位。</p>
<p>依次插入 <code>[1, 6, 11]</code>：</p>
<!-- numbered-subquestions -->
<ol>
<li>三個數最終分別放在哪個索引？</li>
<li>查詢 11 時，會依次檢查哪些索引？</li>
<li>如果刪除 6 時直接把它的位置改成“從未使用的空位”，而查詢遇到空位就停止，
再查詢 11 時會發生什麼？這個查詢結果是否正確？如果有問題，應怎樣避免？</li>
</ol>
<details class="success">
<summary>參考答案</summary>
<ol>
<li>
<p>1 放在索引 1。6 也對映到索引 1，發生衝突後放在索引 2。
11 同樣從索引 1 開始，依次跳過已佔用的索引 1、2，最終放在索引 3。</p>
</li>
<li>
<p>查詢 11 時依次檢查索引 <code>1、2、3</code>，在索引 3 找到它。</p>
</li>
<li>
<p>如果把索引 2 改成表示“從未使用”的空位，查詢 11 時檢查索引 1 後就會在索引 2 停止，
從而錯誤地認為 11 不存在。刪除時應留下“已刪除”標記：
查詢遇到該標記時繼續檢查下一個索引（越過索引 4 後回到索引 0），而以後的插入仍可重新使用這個位置。</p>
</li>
</ol>
</details>
<h2 id="652">6.5.2 &nbsp; 程式設計練習<a class="headerlink" href="#652" title="Permanent link">&para;</a></h2>
<h3 id="1_1">1. &nbsp; 比較兩個字串的字元組成<a class="headerlink" href="#1_1" title="Permanent link">&para;</a></h3>
<p>給定兩個只含小寫英文字母的字串 <code>s</code> 和 <code>t</code>。
可以任意調整 <code>s</code> 中字元的位置，但不能新增、刪除或替換字元。</p>
<p>請判斷調整後能否得到 <code>t</code>；可以則返回 <code>true</code>，否則返回 <code>false</code>。
請使用雜湊表記錄各字母的出現次數，不對字串中的字元排序。</p>
<details class="tip">
<summary>解題提示</summary>
<ol>
<li>兩個字串長度不同，它們的字元組成一定不同</li>
<li>用雜湊表記錄每種字母的數量；掃描 s 時把對應計數加 1</li>
<li>掃描 t 時把對應計數減 1；所有計數最終都為 0，字元組成才相同</li>
</ol>
</details>
<p><a class="rounded-button exercise-button" href="https://leetcode.cn/problems/valid-anagram/" rel="noopener noreferrer" target="_blank">LeetCode</a> <a class="rounded-button exercise-button" href="https://leetcode.cn/problems/valid-anagram/solutions/2362065/242-you-xiao-de-zi-mu-yi-wei-ci-ha-xi-bi-cch7/" rel="noopener noreferrer" target="_blank">題目解析</a></p>
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