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第 4 章 &nbsp; 陣列與鏈結串列
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4.2 &nbsp; 鏈結串列
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第 5 章 &nbsp; 堆疊與佇列
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第 6 章 &nbsp; 雜湊表
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第 7 章 &nbsp; 樹
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第 8 章 &nbsp; 堆積
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<h1 id="136">13.6 &nbsp; 練習<a class="headerlink" href="#136" title="Permanent link">&para;</a></h1>
<h2 id="1361">13.6.1 &nbsp; 知識鞏固<a class="headerlink" href="#1361" title="Permanent link">&para;</a></h2>
<h3 id="1">1. &nbsp; 這段排列演算法會漏掉結果嗎<a class="headerlink" href="#1" title="Permanent link">&para;</a></h3>
<p>一個回溯演算法按 <code>1、2、3</code> 的順序嘗試生成全部排列。每次選擇數字 <code>x</code> 時，它會：</p>
<ol>
<li>把 <code>x</code> 加到當前路徑末尾；</li>
<li>把 <code>x</code> 標記為“已使用”；</li>
<li>遞迴填寫下一個位置。</li>
</ol>
<p>遞迴返回後，同學只從路徑末尾刪除了 <code>x</code>，然後繼續嘗試下一個數字。</p>
<!-- numbered-subquestions -->
<ol>
<li>演算法首先得到哪個排列？它還能得到全部 6 個排列嗎？</li>
<li>遞迴返回上一層前，只刪除路徑末尾的數字是否足夠？如果不夠，還需要做什麼？說明理由。</li>
</ol>
<details class="success">
<summary>參考答案</summary>
<ol>
<li>
<p>它首先得到 <code>[1, 2, 3]</code>，但無法得到全部排列。雖然返回時路徑變短了，
數字 1、2、3 的標記仍都是“已使用”，後續分支便沒有可選數字。</p>
</li>
<li>
<p>不夠。刪除路徑末尾的 <code>x</code> 後，還必須把 <code>x</code> 重新標記為“未使用”。
當前路徑和已使用標記共同描述搜尋狀態；選擇時修改了兩處，回退時也必須把兩處都恢復，
其他分支才能再次選擇 <code>x</code>。</p>
</li>
</ol>
</details>
<h3 id="2">2. &nbsp; 數字的選擇順序重要嗎<a class="headerlink" href="#2" title="Permanent link">&para;</a></h3>
<p>給定排好序的陣列 <code>[2, 3, 5]</code> 和目標值 5，每個數可以重複選擇。
演算法規定每條搜尋路徑中的數字只能按從小到大的順序出現。</p>
<!-- numbered-subquestions -->
<ol>
<li>能得到哪些不同的組合？</li>
<li>為什麼同一組數字不需要按不同順序重複搜尋？“從小到大”的限制起到了什麼作用？</li>
<li>當前路徑為 <code>[3]</code>、還差 2 時，下一個候選數是 3。為什麼此時可以停止檢查這一層後面的所有候選數？</li>
</ol>
<details class="success">
<summary>參考答案</summary>
<ol>
<li>
<p>不同的組合為 <code>[2, 3]</code> 和 <code>[5]</code>。</p>
</li>
<li>
<p>本題把 <code>[2, 3]</code> 和 <code>[3, 2]</code> 看作同一種組合，數字的選擇順序不計入答案。
規定路徑中的數字從小到大出現，就能在搜尋時直接跳過 <code>[3, 2]</code> 這類重複組合。</p>
</li>
<li>
<p>當前還差 2，而候選數 3 已經大於 2。因為陣列已經排好序，
3 後面的候選數只會更大，也都不可能加入當前組合，所以可以直接結束這一層的檢查。</p>
</li>
</ol>
</details>
<h3 id="3">3. &nbsp; 下一枚皇后可以放在哪些位置<a class="headerlink" href="#3" title="Permanent link">&para;</a></h3>
<p>在一個 <code>4 × 4</code> 棋盤上按行放置皇后，行、列下標都從 0 開始。
目前已經在 <code>(0, 1)</code> 和 <code>(1, 3)</code> 放置了皇后，現在要在第 2 行放置下一個皇后。</p>
<!-- numbered-subquestions -->
<ol>
<li>哪些列會因為“同列”而被排除？</li>
<li>在剩餘列中，哪些位置會因為“同一條對角線”而被排除？</li>
<li>第 2 行還剩哪些位置可以嘗試？</li>
</ol>
<details class="success">
<summary>參考答案</summary>
<ol>
<li>
<p>第 1 列和第 3 列已經有皇后，因此位置 <code>(2, 1)</code> 和 <code>(2, 3)</code> 被排除。</p>
</li>
<li>
<p>在剩餘位置中，<code>(2, 2)</code> 與 <code>(1, 3)</code> 位於同一條對角線上，因此也被排除。
位置 <code>(2, 0)</code> 與已有兩個皇后都不在同列或同一條對角線上。</p>
</li>
<li>
<p>第 2 行可以嘗試的位置只有 <code>(2, 0)</code>。</p>
<p>這一步只說明當前放置合法；之後若無法完成棋盤，仍需回退並嘗試更早的其他選擇。</p>
</li>
</ol>
</details>
<h2 id="1362">13.6.2 &nbsp; 程式設計練習<a class="headerlink" href="#1362" title="Permanent link">&para;</a></h2>
<h3 id="1_1">1. &nbsp; 無重複元素的全排列<a class="headerlink" href="#1_1" title="Permanent link">&para;</a></h3>
<p>整數陣列 <code>nums</code> 至少包含一個元素，並且其中各元素互不相同。
請列出把這些元素各使用一次所能形成的全部順序，並將每一種順序作為一個陣列返回。
結果中各排列的先後次序不作要求。
請使用回溯，並用布林陣列記錄每個位置的元素是否已經選入當前排列。</p>
<details class="tip">
<summary>解題提示</summary>
<ol>
<li>遞迴深度表示正在填寫排列中的第幾個位置</li>
<li>每一層只嘗試尚未使用的元素</li>
<li>路徑長度達到 <code>nums</code> 的長度時，把它的副本加入答案</li>
</ol>
</details>
<p><a class="rounded-button exercise-button" href="https://leetcode.cn/problems/permutations/" rel="noopener noreferrer" target="_blank">LeetCode</a> <a class="rounded-button exercise-button" href="https://leetcode.cn/problems/permutations/solutions/2363882/46-quan-pai-lie-hui-su-qing-xi-tu-jie-by-6o7h/" rel="noopener noreferrer" target="_blank">題目解析</a></p>
<blockquote>
<p><strong>說明：</strong> 連結題解透過交換陣列元素，把已選元素依次放到陣列前部；本練習使用布林陣列記錄每個元素是否已選。兩種方法都能避免同一元素被重複選擇，但程式碼結構不同</p>
</blockquote>
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