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"question": {
"questionId": "3471",
"questionFrontendId": "3194",
"categoryTitle": "Algorithms",
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"title": "Minimum Average of Smallest and Largest Elements",
"titleSlug": "minimum-average-of-smallest-and-largest-elements",
"content": "<p>You have an array of floating point numbers <code>averages</code> which is initially empty. You are given an array <code>nums</code> of <code>n</code> integers where <code>n</code> is even.</p>\n\n<p>You repeat the following procedure <code>n / 2</code> times:</p>\n\n<ul>\n\t<li>Remove the <strong>smallest</strong> element, <code>minElement</code>, and the <strong>largest</strong> element <code>maxElement</code>,&nbsp;from <code>nums</code>.</li>\n\t<li>Add <code>(minElement + maxElement) / 2</code> to <code>averages</code>.</li>\n</ul>\n\n<p>Return the <strong>minimum</strong> element in <code>averages</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">nums = [7,8,3,4,15,13,4,1]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">5.5</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<table>\n\t<tbody>\n\t\t<tr>\n\t\t\t<th>step</th>\n\t\t\t<th>nums</th>\n\t\t\t<th>averages</th>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>0</td>\n\t\t\t<td>[7,8,3,4,15,13,4,1]</td>\n\t\t\t<td>[]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>1</td>\n\t\t\t<td>[7,8,3,4,13,4]</td>\n\t\t\t<td>[8]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>2</td>\n\t\t\t<td>[7,8,4,4]</td>\n\t\t\t<td>[8,8]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>3</td>\n\t\t\t<td>[7,4]</td>\n\t\t\t<td>[8,8,6]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>4</td>\n\t\t\t<td>[]</td>\n\t\t\t<td>[8,8,6,5.5]</td>\n\t\t</tr>\n\t</tbody>\n</table>\nThe smallest element of averages, 5.5, is returned.</div>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">nums = [1,9,8,3,10,5]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">5.5</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<table>\n\t<tbody>\n\t\t<tr>\n\t\t\t<th>step</th>\n\t\t\t<th>nums</th>\n\t\t\t<th>averages</th>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>0</td>\n\t\t\t<td><span class=\"example-io\">[1,9,8,3,10,5]</span></td>\n\t\t\t<td>[]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>1</td>\n\t\t\t<td><span class=\"example-io\">[9,8,3,5]</span></td>\n\t\t\t<td>[5.5]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>2</td>\n\t\t\t<td><span class=\"example-io\">[8,5]</span></td>\n\t\t\t<td>[5.5,6]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>3</td>\n\t\t\t<td>[]</td>\n\t\t\t<td>[5.5,6,6.5]</td>\n\t\t</tr>\n\t</tbody>\n</table>\n</div>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">nums = [1,2,3,7,8,9]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">5.0</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<table>\n\t<tbody>\n\t\t<tr>\n\t\t\t<th>step</th>\n\t\t\t<th>nums</th>\n\t\t\t<th>averages</th>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>0</td>\n\t\t\t<td><span class=\"example-io\">[1,2,3,7,8,9]</span></td>\n\t\t\t<td>[]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>1</td>\n\t\t\t<td><span class=\"example-io\">[2,3,7,8]</span></td>\n\t\t\t<td>[5]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>2</td>\n\t\t\t<td><span class=\"example-io\">[3,7]</span></td>\n\t\t\t<td>[5,5]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>3</td>\n\t\t\t<td><span class=\"example-io\">[]</span></td>\n\t\t\t<td>[5,5,5]</td>\n\t\t</tr>\n\t</tbody>\n</table>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= n == nums.length &lt;= 50</code></li>\n\t<li><code>n</code> is even.</li>\n\t<li><code>1 &lt;= nums[i] &lt;= 50</code></li>\n</ul>\n",
"translatedTitle": "最小元素和最大元素的最小平均值",
"translatedContent": "<p>你有一个初始为空的浮点数数组 <code>averages</code>。另给你一个包含 <code>n</code> 个整数的数组 <code>nums</code>,其中 <code>n</code> 为偶数。</p>\n\n<p>你需要重复以下步骤 <code>n / 2</code> 次:</p>\n\n<ul>\n\t<li>从 <code>nums</code> 中移除<strong> 最小 </strong>的元素 <code>minElement</code> 和<strong> 最大 </strong>的元素 <code>maxElement</code>。</li>\n\t<li>将 <code>(minElement + maxElement) / 2</code> 加入到 <code>averages</code> 中。</li>\n</ul>\n\n<p>返回 <code>averages</code> 中的 <strong>最小 </strong>元素。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入:</strong> <span class=\"example-io\">nums = [7,8,3,4,15,13,4,1]</span></p>\n\n<p><strong>输出:</strong> <span class=\"example-io\">5.5</span></p>\n\n<p><strong>解释:</strong></p>\n\n<table>\n\t<tbody>\n\t\t<tr>\n\t\t\t<th>步骤</th>\n\t\t\t<th>nums</th>\n\t\t\t<th>averages</th>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>0</td>\n\t\t\t<td>[7,8,3,4,15,13,4,1]</td>\n\t\t\t<td>[]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>1</td>\n\t\t\t<td>[7,8,3,4,13,4]</td>\n\t\t\t<td>[8]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>2</td>\n\t\t\t<td>[7,8,4,4]</td>\n\t\t\t<td>[8,8]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>3</td>\n\t\t\t<td>[7,4]</td>\n\t\t\t<td>[8,8,6]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>4</td>\n\t\t\t<td>[]</td>\n\t\t\t<td>[8,8,6,5.5]</td>\n\t\t</tr>\n\t</tbody>\n</table>\n返回 averages 中最小的元素,即 5.5。</div>\n\n<p><strong class=\"example\">示例 2</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入:</strong> <span class=\"example-io\">nums = [1,9,8,3,10,5]</span></p>\n\n<p><strong>输出:</strong> <span class=\"example-io\">5.5</span></p>\n\n<p><strong>解释:</strong></p>\n\n<table>\n\t<tbody>\n\t\t<tr>\n\t\t\t<th>步骤</th>\n\t\t\t<th>nums</th>\n\t\t\t<th>averages</th>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>0</td>\n\t\t\t<td>[1,9,8,3,10,5]</td>\n\t\t\t<td>[]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>1</td>\n\t\t\t<td>[9,8,3,5]</td>\n\t\t\t<td>[5.5]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>2</td>\n\t\t\t<td>[8,5]</td>\n\t\t\t<td>[5.5,6]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>3</td>\n\t\t\t<td>[]</td>\n\t\t\t<td>[5.5,6,6.5]</td>\n\t\t</tr>\n\t</tbody>\n</table>\n</div>\n\n<p><strong class=\"example\">示例 3</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入:</strong> <span class=\"example-io\">nums = [1,2,3,7,8,9]</span></p>\n\n<p><strong>输出:</strong> <span class=\"example-io\">5.0</span></p>\n\n<p><strong>解释:</strong></p>\n\n<table>\n\t<tbody>\n\t\t<tr>\n\t\t\t<th>步骤</th>\n\t\t\t<th>nums</th>\n\t\t\t<th>averages</th>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>0</td>\n\t\t\t<td>[1,2,3,7,8,9]</td>\n\t\t\t<td>[]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>1</td>\n\t\t\t<td>[2,3,7,8]</td>\n\t\t\t<td>[5]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>2</td>\n\t\t\t<td>[3,7]</td>\n\t\t\t<td>[5,5]</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td>3</td>\n\t\t\t<td>[]</td>\n\t\t\t<td>[5,5,5]</td>\n\t\t</tr>\n\t</tbody>\n</table>\n</div>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= n == nums.length &lt;= 50</code></li>\n\t<li><code>n</code> 为偶数。</li>\n\t<li><code>1 &lt;= nums[i] &lt;= 50</code></li>\n</ul>\n",
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<code>out-of-bounds<\\/code> \\u548c <code>use-after-free<\\/code> \\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n<p>\\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n<p>\\u5982\\u60f3\\u4f7f\\u7528\\u54c8\\u5e0c\\u8868\\u8fd0\\u7b97, \\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 <a href=\\\"https:\\/\\/troydhanson.github.io\\/uthash\\/\\\" target=\\\"_blank\\\">uthash<\\/a>\\u3002 \\\"uthash.h\\\"\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5bfc\\u5165\\u3002\\u8bf7\\u770b\\u5982\\u4e0b\\u793a\\u4f8b:<\\/p>\\r\\n\\r\\n<p><b>1. \\u5f80\\u54c8\\u5e0c\\u8868\\u4e2d\\u6dfb\\u52a0\\u4e00\\u4e2a\\u5bf9\\u8c61\\uff1a<\\/b>\\r\\n<pre>\\r\\nstruct hash_entry {\\r\\n int id; \\/* we'll use this field as the key *\\/\\r\\n char name[10];\\r\\n UT_hash_handle hh; \\/* makes this structure hashable *\\/\\r\\n};\\r\\n\\r\\nstruct hash_entry *users = NULL;\\r\\n\\r\\nvoid 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