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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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第 11 章 &nbsp; 排序
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<h1 id="148">14.8 &nbsp; 練習<a class="headerlink" href="#148" title="Permanent link">&para;</a></h1>
<h2 id="1481">14.8.1 &nbsp; 知識鞏固<a class="headerlink" href="#1481" title="Permanent link">&para;</a></h2>
<h3 id="1">1. &nbsp; 什麼時候適合使用動態規劃<a class="headerlink" href="#1" title="Permanent link">&para;</a></h3>
<p>一位同學說：“只要能寫出遞推式，就應該使用動態規劃。”
請判斷下面三個任務分別更適合使用動態規劃、回溯，還是使用迴圈或數學公式而不建立 <code>dp</code> 表，並寫出一個關鍵理由。</p>
<!-- numbered-subquestions -->
<ol>
<li>使用面值 <code>[1, 3, 4]</code> 的硬幣湊出金額 6，求最少硬幣數，每種硬幣可重複使用；</li>
<li>輸出 <code>[1, 2, 3]</code> 的全部排列；</li>
<li>
<p>計算 <span class="arithmatex">\(1 + 2 + \dots + n\)</span>。</p>
<p>對於你判斷為適合動態規劃的任務，再說明 <code>dp[i]</code> 表示什麼。</p>
</li>
</ol>
<details class="success">
<summary>參考答案</summary>
<ol>
<li>
<p>適合動態規劃。令 <code>dp[i]</code> 表示湊出金額 <code>i</code> 所需的最少硬幣數。
對於每枚不超過 <code>i</code> 的硬幣 <code>c</code>，都可以把 <code>dp[i-c] + 1</code> 作為候選答案，
再從這些候選值中取最小值。不同選擇會反覆遇到同一金額，較大金額的最優解也能由較小金額的最優解構成。
金額 6 的答案是 2，即 <code>3 + 3</code>。</p>
</li>
<li>
<p>應使用回溯。題目要求逐個生成全部 6 個排列，回溯可以系統地嘗試一個選擇、繼續搜尋，
再撤銷選擇並嘗試其他分支。無論採用什麼方法，實際輸出這些排列時都不能跳過列舉。</p>
</li>
<li>
<p>使用迴圈或等差數列公式即可。雖然能寫出 <code>S(i) = S(i-1) + i</code>，但計算 <code>S(i)</code> 時只依賴一個更小的 <code>S(i-1)</code>，
每個部分和只需計算一次，不存在重複子問題，因此無需建立 <code>dp</code> 表。“能寫遞推式”不等於“需要動態規劃”。</p>
</li>
</ol>
</details>
<h3 id="2">2. &nbsp; 背包表中的一個格子怎麼算<a class="headerlink" href="#2" title="Permanent link">&para;</a></h3>
<p>0-1 背包：物品重量 <code>wgt = [1, 2, 3]</code> ，價值 <code>val = [5, 11, 15]</code> ，背包容量 4 。
<code>dp[i][c]</code> 表示只考慮前 <span class="arithmatex">\(i\)</span> 件物品、背包容量上限為 <span class="arithmatex">\(c\)</span> 時，能夠取得的最大價值；
不要求恰好裝滿背包。</p>
<p>現在只計算狀態 <code>dp[3][4]</code>。已知 <code>dp[2][4] = 16</code>、<code>dp[2][1] = 5</code>：</p>
<!-- numbered-subquestions -->
<ol>
<li>不選第 3 件物品時，候選價值是多少？</li>
<li>選擇第 3 件物品時，還剩多少容量，候選價值是多少？</li>
<li><code>dp[3][4]</code> 應取多少？對應選擇了哪些物品？</li>
</ol>
<details class="success">
<summary>參考答案</summary>
<ol>
<li>
<p>不選第 3 件物品時，沿用前兩件物品的結果，候選價值為 <code>dp[2][4] = 16</code>。</p>
</li>
<li>
<p>第 3 件物品重量為 3，放入後還剩容量 <span class="arithmatex">\(4-3=1\)</span>；候選價值為
<code>dp[2][1] + 15 = 5 + 15 = 20</code>。</p>
</li>
<li>
<p>比較 16 和 20，得到 <code>dp[3][4] = 20</code>。它對應選擇第 1、3 件物品，
總重量為 <span class="arithmatex">\(1+3=4\)</span>，總價值為 <span class="arithmatex">\(5+15=20\)</span>。</p>
<p>這個狀態的計算體現了 0-1 背包的一次“選或不選”比較。</p>
</li>
</ol>
</details>
<h3 id="3">3. &nbsp; 背包容量應該按什麼順序更新<a class="headerlink" href="#3" title="Permanent link">&para;</a></h3>
<p>一個 0-1 背包只有一件物品：重量為 2，價值為 5；背包容量為 4。
每件物品最多選擇一次，初始一維陣列為 <code>dp = [0, 0, 0, 0, 0]</code>。</p>
<p>一位同學在處理這件物品時，按容量從 2 到 4 更新：</p>
<ul>
<li>更新 <code>dp[2]</code> 後得到 5；</li>
<li>更新 <code>dp[3]</code> 後也得到 5；</li>
<li>更新 <code>dp[4]</code> 時又使用剛得到的 <code>dp[2]</code>，於是得到 <code>dp[4] = 10</code>。</li>
</ul>
<!-- numbered-subquestions -->
<ol>
<li><code>dp[4] = 10</code> 這個結果是否正確？為什麼？</li>
<li>根據“每件物品最多選擇一次”的條件，<code>dp[4]</code> 應為多少？</li>
<li>處理每件物品時，容量應從大到小還是從小到大更新？這樣做避免了什麼問題？</li>
</ol>
<details class="success">
<summary>參考答案</summary>
<ol>
<li>
<p>這個結果不正確。價值 10 相當於把這件價值為 5 的物品放入了兩次，
違反了“每件物品最多選擇一次”的條件。</p>
</li>
<li>
<p>背包中最多隻能放入這一件物品，所以正確的 <code>dp[4]</code> 為 5。</p>
</li>
<li>
<p>容量應從大到小更新，即依次處理 4、3、2。
這樣在計算 <code>dp[c]</code> 時，所讀取的 <code>dp[c-2]</code> 仍是處理當前物品之前的結果，
不會在同一輪中重複使用當前物品。</p>
</li>
</ol>
</details>
<h2 id="1482">14.8.2 &nbsp; 程式設計練習<a class="headerlink" href="#1482" title="Permanent link">&para;</a></h2>
<h3 id="1_1">1. &nbsp; 爬樓梯的方案數<a class="headerlink" href="#1_1" title="Permanent link">&para;</a></h3>
<p>一段樓梯共有 <code>n</code> 階。你每次只能向上走 1 階或 2 階，且必須恰好到達第 <code>n</code> 階。
請計算共有多少種不同的走法。規定 <code>n &gt;= 1</code>，走法只按每一步走 1 階還是 2 階來區分。
請使用一維動態規劃陣列完成，暫不使用只保留兩個狀態的空間最佳化寫法。</p>
<details class="tip">
<summary>解題提示</summary>
<ol>
<li>到達第 i 階的最後一步，只可能跨 1 階或 2 階</li>
<li>因此有 dp[i] = dp[i-1] + dp[i-2]</li>
<li>先處理 n 為 1 和 2 的情況，再從第 3 階開始填表</li>
</ol>
</details>
<p><a class="rounded-button exercise-button" href="https://leetcode.cn/problems/climbing-stairs/" rel="noopener noreferrer" target="_blank">LeetCode</a> <a class="rounded-button exercise-button" href="https://leetcode.cn/problems/climbing-stairs/solutions/2361764/70-pa-lou-ti-dong-tai-gui-hua-qing-xi-tu-ruwa/" rel="noopener noreferrer" target="_blank">題目解析</a></p>
<blockquote>
<p><strong>說明：</strong> 題解講解了一維 dp 狀態和轉移，但給出的程式碼採用滾動變數壓縮空間；本練習要求先實現一維 dp 表，滾動變數僅作為選做最佳化</p>
</blockquote>
<h3 id="2-0-1">2. &nbsp; 0-1 背包<a class="headerlink" href="#2-0-1" title="Permanent link">&para;</a></h3>
<p>給定等長陣列 <code>wgt</code> 和 <code>val</code>，第 <code>i</code> 件物品的重量為正整數 <code>wgt[i]</code>、價值為非負整數 <code>val[i]</code>，
背包容量 <code>cap</code> 為非負整數。每件物品最多選擇一次，請在總重量不超過 <code>cap</code> 的條件下，
求能夠裝入背包的最大總價值。請使用一維動態規劃實現。</p>
<details class="tip">
<summary>解題提示</summary>
<ol>
<li>初始化長度為 cap + 1 的陣列 dp，其中 dp[c] 表示容量上限為 c 時的最大價值</li>
<li>處理物品 i 時，比較“不選它”的 dp[c] 與“選它”的 dp[c-wgt[i]] + val[i]</li>
<li>容量必須從大到小更新，避免在同一輪中重複選擇當前物品</li>
</ol>
</details>
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