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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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<h1 id="26">2.6 &nbsp; 練習<a class="headerlink" href="#26" title="Permanent link">&para;</a></h1>
<h2 id="261">2.6.1 &nbsp; 知識鞏固<a class="headerlink" href="#261" title="Permanent link">&para;</a></h2>
<h3 id="1">1. &nbsp; 迭代與遞迴的時間和空間<a class="headerlink" href="#1" title="Permanent link">&para;</a></h3>
<p>下面兩段程式碼都計算 <span class="arithmatex">\(1 + 2 + \dots + n\)</span>（設 <span class="arithmatex">\(n \ge 1\)</span>）。請把 <code>n</code> 設為 4，
按照程式實際執行的順序回答問題，然後比較兩種寫法的效率。</p>
<div class="highlight"><pre><span></span><code><a id="__codelineno-0-1" name="__codelineno-0-1" href="#__codelineno-0-1"></a><span class="k">def</span><span class="w"> </span><span class="nf">sum_iter</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
<a id="__codelineno-0-2" name="__codelineno-0-2" href="#__codelineno-0-2"></a> <span class="n">s</span> <span class="o">=</span> <span class="mi">0</span>
<a id="__codelineno-0-3" name="__codelineno-0-3" href="#__codelineno-0-3"></a> <span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="mi">1</span><span class="p">,</span> <span class="n">n</span> <span class="o">+</span> <span class="mi">1</span><span class="p">):</span>
<a id="__codelineno-0-4" name="__codelineno-0-4" href="#__codelineno-0-4"></a> <span class="n">s</span> <span class="o">+=</span> <span class="n">i</span>
<a id="__codelineno-0-5" name="__codelineno-0-5" href="#__codelineno-0-5"></a> <span class="k">return</span> <span class="n">s</span>
<a id="__codelineno-0-6" name="__codelineno-0-6" href="#__codelineno-0-6"></a>
<a id="__codelineno-0-7" name="__codelineno-0-7" href="#__codelineno-0-7"></a><span class="k">def</span><span class="w"> </span><span class="nf">sum_recur</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
<a id="__codelineno-0-8" name="__codelineno-0-8" href="#__codelineno-0-8"></a> <span class="k">if</span> <span class="n">n</span> <span class="o">==</span> <span class="mi">1</span><span class="p">:</span>
<a id="__codelineno-0-9" name="__codelineno-0-9" href="#__codelineno-0-9"></a> <span class="k">return</span> <span class="mi">1</span>
<a id="__codelineno-0-10" name="__codelineno-0-10" href="#__codelineno-0-10"></a> <span class="k">return</span> <span class="n">n</span> <span class="o">+</span> <span class="n">sum_recur</span><span class="p">(</span><span class="n">n</span> <span class="o">-</span> <span class="mi">1</span><span class="p">)</span>
</code></pre></div>
<!-- numbered-subquestions -->
<ol>
<li>執行 <code>sum_iter(4)</code> 時，每輪迴圈結束後，變數 <code>s</code> 的值分別是多少？</li>
<li>執行 <code>sum_recur(4)</code> 時，會依次呼叫哪些函式？從最深的一層開始返回時，結果怎樣得到？</li>
<li>兩種寫法的時間複雜度和空間複雜度分別是多少？結合第 1、2 問的執行過程說明理由。</li>
</ol>
<details class="success">
<summary>參考答案</summary>
<ol>
<li>
<p>迴圈變數 <code>i</code> 依次為 <code>1、2、3、4</code>，每輪結束後，<code>s</code> 依次變為
<code>1、3、6、10</code>，所以 <code>sum_iter(4)</code> 返回 10。</p>
</li>
<li>
<p>函式依次呼叫
<code>sum_recur(4) → sum_recur(3) → sum_recur(2) → sum_recur(1)</code>。
<code>sum_recur(1)</code> 返回 1，隨後各層依次得到 <code>2 + 1 = 3</code>、<code>3 + 3 = 6</code>、<code>4 + 6 = 10</code>。
在最深處，4 次函式呼叫都尚未結束。</p>
</li>
<li>
<p>兩段程式碼都進行與 <span class="arithmatex">\(n\)</span> 成正比的迴圈或呼叫，因此時間複雜度均為 <span class="arithmatex">\(O(n)\)</span> 。
空間複雜度不同：迭代版只使用常數個變數，為 <span class="arithmatex">\(O(1)\)</span> ；
遞迴版在到達終止條件前，前面的函式呼叫都要等待返回結果，因此呼叫堆疊中最多同時儲存 <span class="arithmatex">\(n\)</span> 次呼叫，
空間複雜度為 <span class="arithmatex">\(O(n)\)</span>。</p>
<p>分析空間複雜度時，除程式碼中的變數外，還要考慮遞迴呼叫佔用的空間。</p>
</li>
</ol>
</details>
<h3 id="2">2. &nbsp; 三段程式碼的時間複雜度<a class="headerlink" href="#2" title="Permanent link">&para;</a></h3>
<p>以下三個程式碼片段的輸入均為正整數 <span class="arithmatex">\(n\)</span> 。請按時間複雜度從低到高排序，並寫出各自的複雜度。</p>
<div class="highlight"><pre><span></span><code><a id="__codelineno-1-1" name="__codelineno-1-1" href="#__codelineno-1-1"></a><span class="c1"># 片段一</span>
<a id="__codelineno-1-2" name="__codelineno-1-2" href="#__codelineno-1-2"></a><span class="n">s</span> <span class="o">=</span> <span class="mi">0</span>
<a id="__codelineno-1-3" name="__codelineno-1-3" href="#__codelineno-1-3"></a><span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
<a id="__codelineno-1-4" name="__codelineno-1-4" href="#__codelineno-1-4"></a> <span class="n">s</span> <span class="o">+=</span> <span class="n">i</span>
<a id="__codelineno-1-5" name="__codelineno-1-5" href="#__codelineno-1-5"></a>
<a id="__codelineno-1-6" name="__codelineno-1-6" href="#__codelineno-1-6"></a><span class="c1"># 片段二</span>
<a id="__codelineno-1-7" name="__codelineno-1-7" href="#__codelineno-1-7"></a><span class="n">s</span> <span class="o">=</span> <span class="mi">0</span>
<a id="__codelineno-1-8" name="__codelineno-1-8" href="#__codelineno-1-8"></a><span class="k">for</span> <span class="n">i</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">n</span><span class="p">):</span>
<a id="__codelineno-1-9" name="__codelineno-1-9" href="#__codelineno-1-9"></a> <span class="k">for</span> <span class="n">j</span> <span class="ow">in</span> <span class="nb">range</span><span class="p">(</span><span class="n">i</span><span class="p">,</span> <span class="n">n</span><span class="p">):</span>
<a id="__codelineno-1-10" name="__codelineno-1-10" href="#__codelineno-1-10"></a> <span class="n">s</span> <span class="o">+=</span> <span class="n">j</span>
<a id="__codelineno-1-11" name="__codelineno-1-11" href="#__codelineno-1-11"></a>
<a id="__codelineno-1-12" name="__codelineno-1-12" href="#__codelineno-1-12"></a><span class="c1"># 片段三</span>
<a id="__codelineno-1-13" name="__codelineno-1-13" href="#__codelineno-1-13"></a><span class="k">while</span> <span class="n">n</span> <span class="o">&gt;</span> <span class="mi">1</span><span class="p">:</span>
<a id="__codelineno-1-14" name="__codelineno-1-14" href="#__codelineno-1-14"></a> <span class="n">n</span> <span class="o">=</span> <span class="n">n</span> <span class="o">//</span> <span class="mi">2</span>
</code></pre></div>
<details class="success">
<summary>參考答案</summary>
<p>從低到高為：片段三 <span class="arithmatex">\(O(\log n)\)</span>、片段一 <span class="arithmatex">\(O(n)\)</span>、片段二 <span class="arithmatex">\(O(n^2)\)</span>。
片段三每輪把 <span class="arithmatex">\(n\)</span> 縮小為原來的一半，約迴圈 <span class="arithmatex">\(\log_2 n\)</span> 次。
片段一的迴圈恰好執行 <span class="arithmatex">\(n\)</span> 次。片段二的內層迴圈次數依次為
<span class="arithmatex">\(n,n-1,\dots,1\)</span>，總次數為 <span class="arithmatex">\(n(n+1)/2\)</span>，因此屬於平方階。</p>
</details>
<h3 id="3">3. &nbsp; 哪種反轉更節省空間<a class="headerlink" href="#3" title="Permanent link">&para;</a></h3>
<p>要將陣列 <code>nums</code> 中的元素全部反轉，有兩種做法：</p>
<!-- numbered-subquestions -->
<ol>
<li>新建一個等長陣列 <code>res</code>，倒序複製後返回；</li>
<li>
<p>用兩個索引 <code>i</code> 和 <code>j</code> 分別從首、尾向中間移動，逐對交換 <code>nums[i]</code> 與 <code>nums[j]</code> 。</p>
<p>兩種做法的空間複雜度各是多少？哪種屬於“原地”操作？</p>
</li>
</ol>
<details class="success">
<summary>參考答案</summary>
<ol>
<li>
<p>需要與輸入等長的輔助陣列，空間複雜度 <span class="arithmatex">\(O(n)\)</span>。</p>
</li>
<li>
<p>只使用兩個索引變數，
空間複雜度 <span class="arithmatex">\(O(1)\)</span> ，屬於原地操作。</p>
<p>需要注意：原地反轉會修改輸入陣列，
僅在允許修改輸入時才應優先選用；若需保留原陣列，第 1 種做法的複製開銷不可避免。</p>
</li>
</ol>
</details>
<h2 id="262">2.6.2 &nbsp; 程式設計練習<a class="headerlink" href="#262" title="Permanent link">&para;</a></h2>
<h3 id="1_1">1. &nbsp; 費波那契數<a class="headerlink" href="#1_1" title="Permanent link">&para;</a></h3>
<p>費波那契數列滿足：<span class="arithmatex">\(F(0)=0\)</span>、<span class="arithmatex">\(F(1)=1\)</span>，並且當 <span class="arithmatex">\(n\ge2\)</span> 時，
<span class="arithmatex">\(F(n)=F(n-1)+F(n-2)\)</span>。</p>
<p>給定非負整數 <code>n</code>，請使用迴圈計算並返回 <span class="arithmatex">\(F(n)\)</span>，不使用遞迴。</p>
<details class="tip">
<summary>解題提示</summary>
<ol>
<li>先單獨處理 n 為 0 和 1 的情況</li>
<li>計算下一項時只需要前兩項，無須儲存整個數列</li>
<li>更新兩個變數時，注意不要過早覆蓋仍會用到的舊值</li>
</ol>
</details>
<p><a class="rounded-button exercise-button" href="https://leetcode.cn/problems/fibonacci-number/" rel="noopener noreferrer" target="_blank">LeetCode</a> <a class="rounded-button exercise-button" href="https://leetcode.cn/problems/fibonacci-number/solutions/2361746/509-fei-bo-na-qi-shu-dong-tai-gui-hua-qi-so8h/" rel="noopener noreferrer" target="_blank">題目解析</a></p>
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