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第 10 章 &nbsp; 探索
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<h1 id="107">10.7 &nbsp; 練習問題<a class="headerlink" href="#107" title="この見出しへのリンク">&para;</a></h1>
<h2 id="1071">10.7.1 &nbsp; 知識の確認<a class="headerlink" href="#1071" title="この見出しへのリンク">&para;</a></h2>
<h3 id="1">1. &nbsp; 二分探索で探索区間を狭めるには<a class="headerlink" href="#1" title="この見出しへのリンク">&para;</a></h3>
<p>整列済み配列 <code>[2, 5, 8, 12, 16, 23, 38]</code> から 16 を探します。
両端を含む閉区間 <code>[i, j]</code> を使い、中点は
<span class="arithmatex">\(m=i+(j-i)/2\)</span>(切り捨て)とします。</p>
<p>見つかるまで、各反復の <code>(i, j, m)</code>、中点の要素、
次に区間をどう狭めるかを書いてください。</p>
<details class="success">
<summary>解答例</summary>
<p>各反復の探索過程は、次のとおりです。</p>
<table>
<thead>
<tr>
<th>反復</th>
<th><code>(i, j, m)</code></th>
<th>中点の要素</th>
<th>次の操作</th>
</tr>
</thead>
<tbody>
<tr>
<td>1</td>
<td><code>(0, 6, 3)</code></td>
<td>12</td>
<td><code>12 &lt; 16</code> なので <code>i = 4</code> とします</td>
</tr>
<tr>
<td>2</td>
<td><code>(4, 6, 5)</code></td>
<td>23</td>
<td><code>23 &gt; 16</code> なので <code>j = 4</code> とします</td>
</tr>
<tr>
<td>3</td>
<td><code>(4, 4, 4)</code></td>
<td>16</td>
<td>見つかったので、インデックス 4 を返します</td>
</tr>
</tbody>
</table>
<p>配列は整列済みなので、中点の値が目的の値より小さければ、中点とその左側を除外できます。
中点の値が目的の値より大きければ、中点とその右側を除外できます。</p>
</details>
<h3 id="2">2. &nbsp; 重複する要素の左右の境界<a class="headerlink" href="#2" title="この見出しへのリンク">&para;</a></h3>
<p>配列 <code>[1, 2, 2, 2, 4, 6]</code> から 2 を探します。
ある生徒が二分探索でインデックス 2 に目的の値を見つけ、すぐに返して、
「インデックス 2 が、2 の左境界だ」と言いました。</p>
<!-- numbered-subquestions -->
<ol>
<li>この生徒の発言は正しいですか。2 の左右の境界はそれぞれどこですか。理由も説明してください。</li>
<li>左境界を探す場合、中点の要素が目的の値と等しかったら、次はどちら側を探し続けますか。</li>
<li>右境界を探す場合は、どちら側を探し続けますか。方向だけを説明し、探索過程全体を書く必要はありません。</li>
</ol>
<details class="success">
<summary>解答例</summary>
<ol>
<li>
<p>この生徒の発言は正しくありません。一つの 2 を見つけてすぐに返すと、どれか一つの 2 を見つけたことしか保証できず、最も左や右の 2 を見つけたとは限りません。
この問題の左境界はインデックス 1、右境界はインデックス 3 です。</p>
</li>
<li>
<p>左境界を探すときは、中点の要素が 2 と等しくても、左側を探し続けます。
例えば、両端を含む閉区間を使う場合は <code>j = m - 1</code> とします。</p>
</li>
<li>
<p>右境界を探すときは、中点の要素が 2 と等しかったら、右側を探し続けます。
例えば <code>i = m + 1</code> とします。</p>
</li>
</ol>
</details>
<h3 id="3">3. &nbsp; データに合った探索方法を選ぶ<a class="headerlink" href="#3" title="この見出しへのリンク">&para;</a></h3>
<p>次の各場面に適した方法を「線形探索、二分探索、ハッシュテーブル」から選び、理由を説明してください。</p>
<!-- numbered-subquestions -->
<ol>
<li>整列済みで今後変化しない <span class="arithmatex">\(10^7\)</span> 個の整数を繰り返し探索します。ほかのデータ構造は追加で作りません。</li>
<li>挿入と削除が頻繁なデータ集合で、キーの有無を繰り返し調べます。整列状態を保つ必要はなく、範囲探索もしません。</li>
<li>未整列配列から、ある値を一回だけ探します。</li>
</ol>
<details class="success">
<summary>解答例</summary>
<ol>
<li>二分探索です。データが整列済みで変化せず、<span class="arithmatex">\(O(\log n)\)</span> 時間で探索でき、追加の領域も不要です。</li>
<li>ハッシュテーブルです。ハッシュ関数がキーを各バケットへ比較的均等に分散できる場合、挿入、削除、キーによる探索の平均時間計算量は、いずれも <span class="arithmatex">\(O(1)\)</span> です。</li>
<li>
<p>先頭から末尾まで直接走査します。一回しか探さない場合、ソートやハッシュテーブルの構築では、先に配列全体を処理する必要があり、
その一回の作業の総量は減りません。</p>
<p>選択は、データが整列済みか、追加の構造を作れるか、照会回数はいくつか、どの操作が必要かによって決まります。</p>
</li>
</ol>
</details>
<h2 id="1072">10.7.2 &nbsp; プログラミング演習<a class="headerlink" href="#1072" title="この見出しへのリンク">&para;</a></h2>
<h3 id="1_1">1. &nbsp; 整列済み配列の二分探索<a class="headerlink" href="#1_1" title="この見出しへのリンク">&para;</a></h3>
<p>狭義の昇順に並んだ整数配列 <code>nums</code> と目的の値 <code>target</code> が与えられます。二分探索で <code>target</code> を探し、存在すれば配列内のインデックスを、存在しなければ -1 を返してください。</p>
<details class="tip">
<summary>解くためのヒント</summary>
<ol>
<li>初期区間は left = 0、right = n - 1 で、区間が空でない条件は left &lt;= right です</li>
<li>mid = left + (right - left) // 2 で中点を計算します</li>
<li><code>nums[mid]</code> が <code>target</code> より小さければ左境界を <code>mid + 1</code> へ、<code>nums[mid]</code> が <code>target</code> より大きければ右境界を <code>mid - 1</code> へ移します。等しければ、すぐに返します</li>
</ol>
</details>
<p><a class="rounded-button exercise-button" href="https://leetcode.com/problems/binary-search/" rel="noopener noreferrer" target="_blank">LeetCode</a></p>
<h3 id="2_1">2. &nbsp; 整列済み配列への挿入位置<a class="headerlink" href="#2_1" title="この見出しへのリンク">&para;</a></h3>
<p>狭義の昇順に並んだ整数配列 <code>nums</code> と目的の値 <code>target</code> が与えられます。</p>
<p><code>target</code> が配列にあれば、そのインデックスを返してください。なければ、<code>target</code> を挿入しても狭義の昇順を保てる位置を返してください。答えは 0 の場合も、配列の長さと等しい場合もあります。二分探索を使ってください。</p>
<details class="tip">
<summary>解くためのヒント</summary>
<ol>
<li>答えは 0 の場合も、配列の長さ n の場合もあります</li>
<li>両端を含む閉区間を使う場合、<code>nums[mid]</code> が <code>target</code> 以上なら <code>right = mid - 1</code> として、さらに左を調べます。そうでなければ <code>left = mid + 1</code> とします</li>
<li>ループが終わったときの left が挿入位置です</li>
</ol>
</details>
<p><a class="rounded-button exercise-button" href="https://leetcode.com/problems/search-insert-position/" rel="noopener noreferrer" target="_blank">LeetCode</a></p>
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