leetcode-problemset/leetcode-cn/originData/count-the-number-of-infection-sequences.json

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        "question": {
            "questionId": "3224",
            "questionFrontendId": "2954",
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            "title": "Count the Number of Infection Sequences",
            "titleSlug": "count-the-number-of-infection-sequences",
            "content": "<p>You are given an integer <code>n</code> and an array <code>sick</code> sorted in increasing order, representing positions of infected people in a line of <code>n</code> people.</p>\n\n<p>At each step, <strong>one </strong>uninfected person <strong>adjacent</strong> to an infected person gets infected. This process continues until everyone is infected.</p>\n\n<p>An <strong>infection sequence</strong> is the order in which uninfected people become infected, excluding those initially infected.</p>\n\n<p>Return the number of different infection sequences possible, modulo <code>10<sup>9</sup>+7</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\">n = 5, sick = [0,4]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">4</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>There is a total of 6 different sequences overall.</p>\n\n<ul>\n\t<li>Valid infection sequences are <code>[1,2,3]</code>, <code>[1,3,2]</code>, <code>[3,2,1]</code> and <code>[3,1,2]</code>.</li>\n\t<li><code>[2,3,1]</code> and <code>[2,1,3]</code> are not valid infection sequences because the person at index 2 cannot be infected at the first step.</li>\n</ul>\n</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\">n = 4, sick = [1]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">3</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>There is a total of 6 different sequences overall.</p>\n\n<ul>\n\t<li>Valid infection sequences are <code>[0,2,3]</code>, <code>[2,0,3]</code> and <code>[2,3,0]</code>.</li>\n\t<li><code>[3,2,0]</code>, <code>[3,0,2]</code>, and <code>[0,3,2]</code> are not valid infection sequences because the infection starts at the person at index 1, then the order of infection is 2, then 3, and hence 3 cannot be infected earlier than 2.</li>\n</ul>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= sick.length &lt;= n - 1</code></li>\n\t<li><code>0 &lt;= sick[i] &lt;= n - 1</code></li>\n\t<li><code>sick</code> is sorted in increasing order.</li>\n</ul>\n",
            "translatedTitle": "统计感冒序列的数目",
            "translatedContent": "<p>给你一个整数&nbsp;<code>n</code>&nbsp;和一个下标从 <strong>0</strong>&nbsp;开始的整数数组&nbsp;<code>sick</code>&nbsp;，数组按 <strong>升序</strong>&nbsp;排序。</p>\n\n<p>有&nbsp;<code>n</code>&nbsp;位小朋友站成一排，按顺序编号为 <code>0</code>&nbsp;到 <code>n - 1</code>&nbsp;。数组&nbsp;<code>sick</code>&nbsp;包含一开始得了感冒的小朋友的位置。如果位置为&nbsp;<code>i</code>&nbsp;的小朋友得了感冒，他会传染给下标为 <code>i - 1</code>&nbsp;或者 <code>i + 1</code>&nbsp;的小朋友，<strong>前提</strong> 是被传染的小朋友存在且还没有得感冒。每一秒中， <strong>至多一位</strong>&nbsp;还没感冒的小朋友会被传染。</p>\n\n<p>经过有限的秒数后，队列中所有小朋友都会感冒。<strong>感冒序列</strong>&nbsp;指的是 <strong>所有</strong>&nbsp;一开始没有感冒的小朋友最后得感冒的顺序序列。请你返回所有感冒序列的数目。</p>\n\n<p>由于答案可能很大，请你将答案对&nbsp;<code>10<sup>9</sup> + 7</code>&nbsp;取余后返回。</p>\n\n<p><b>注意</b>，感冒序列 <strong>不</strong> 包含一开始就得了感冒的小朋友的下标。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>n = 5, sick = [0,4]\n<b>输出：</b>4\n<b>解释：</b>一开始，下标为 1 ，2 和 3 的小朋友没有感冒。总共有 4 个可能的感冒序列：\n- 一开始，下标为 1 和 3 的小朋友可以被传染，因为他们分别挨着有感冒的小朋友 0 和 4 ，令下标为 1 的小朋友先被传染。\n然后，下标为 2 的小朋友挨着感冒的小朋友 1 ，下标为 3 的小朋友挨着感冒的小朋友 4 ，两位小朋友都可以被传染，令下标为 2 的小朋友被传染。\n最后，下标为 3 的小朋友被传染，因为他挨着感冒的小朋友 2 和 4 ，感冒序列为 [1,2,3] 。\n- 一开始，下标为 1 和 3 的小朋友可以被传染，因为他们分别挨着感冒的小朋友 0 和 4 ，令下标为 1 的小朋友先被传染。\n然后，下标为 2 的小朋友挨着感冒的小朋友 1 ，下标为 3 的小朋友挨着感冒的小朋友 4 ，两位小朋友都可以被传染，令下标为 3 的小朋友被传染。\n最后，下标为 2 的小朋友被传染，因为他挨着感冒的小朋友 1 和 3 ，感冒序列为  [1,3,2] 。\n- 感冒序列 [3,1,2] ，被传染的顺序：[<strong><em>0</em></strong>,1,2,3,<strong><em>4</em></strong>] =&gt; [<strong><em>0</em></strong>,1,2,<strong><em>3</em></strong>,<strong><em>4</em></strong>] =&gt; [<strong><em>0</em></strong>,<strong><em>1</em></strong>,2,<em><strong>3</strong></em>,<strong><em>4</em></strong>] =&gt; [<strong><em>0</em></strong>,<strong><em>1</em></strong>,<strong><em>2</em></strong>,<strong><em>3</em></strong>,<strong><em>4</em></strong>] 。\n- 感冒序列 [3,2,1] ，被传染的顺序：[<strong><em>0</em></strong>,1,2,3,<strong><em>4</em></strong>] =&gt; [<strong><em>0</em></strong>,1,2,<strong><em>3</em></strong>,<strong><em>4</em></strong>] =&gt; [<strong><em>0</em></strong>,1,<strong><em>2</em></strong>,<strong><em>3</em></strong>,<strong><em>4</em></strong>] =&gt; [<strong><em>0</em></strong>,<strong><em>1</em></strong>,<strong><em>2</em></strong>,<strong><em>3</em></strong>,<strong><em>4</em></strong>] 。\n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>n = 4, sick = [1]\n<b>输出：</b>3\n<b>解释：</b>一开始，下标为 0 ，2 和 3 的小朋友没有感冒。总共有 3 个可能的感冒序列：\n- 感冒序列 [0,2,3] ，被传染的顺序：[0,<strong><em>1</em></strong>,2,3] =&gt; [<strong><em>0</em></strong>,<strong><em>1</em></strong>,2,3] =&gt; [<strong><em>0</em></strong>,<strong><em>1</em></strong>,<strong><em>2</em></strong>,3] =&gt; [<strong><em>0</em></strong>,<strong><em>1</em></strong>,<strong><em>2</em></strong>,<strong><em>3</em></strong>] 。\n- 感冒序列 [2,0,3] ，被传染的顺序：[0,<strong><em>1</em></strong>,2,3] =&gt; [0,<strong><em>1</em></strong>,<strong><em>2</em></strong>,3] =&gt; [<strong><em>0</em></strong>,<strong><em>1</em></strong>,<strong><em>2</em></strong>,3] =&gt; [<strong><em>0</em></strong>,<strong><em>1</em></strong>,<strong><em>2</em></strong>,<strong><em>3</em></strong>] 。\n- 感冒序列 [2,3,0] ，被传染的顺序：[0,<strong><em>1</em></strong>,2,3] =&gt; [0,<strong><em>1</em></strong>,<strong><em>2</em></strong>,3] =&gt; [0,<strong><em>1</em></strong>,<strong><em>2</em></strong>,<strong><em>3</em></strong>] =&gt; [<strong><em>0</em></strong>,<strong><em>1</em></strong>,<strong><em>2</em></strong>,<strong><em>3</em></strong>] 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= sick.length &lt;= n - 1</code></li>\n\t<li><code>0 &lt;= sick[i] &lt;= n - 1</code></li>\n\t<li><code>sick</code>&nbsp;按升序排列。</li>\n</ul>\n",
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                "Consider infected children as <code>0</code> and non-infected as <code>1</code>, then divide the array into segments with the same value.",
                "For each segment of non-infected children whose indices are <code>[i, j]</code> and indices <code>(i - 1)</code> and <code>(j + 1)</code>, if they exist, are already infected. Then if <code>i == 0</code> or <code>j == n - 1</code>, each second there is only one kid that can be infected (which is at the other endpoint).",
                "If <code>i > 0</code> and <code>j < n - 1</code>, we have two choices per second since the children at the two endpoints can both be the infect candidates. So there are <code>2<sup>j - i</sup></code> orders to infect all children in the segment.",
                "Each second we can select a segment and select one endpoint from it.",
                "The answer is: \r\n<code>S! / (len[1]! * len[2]! * ... * len[m]! * len<sub>start</sub>! * len<sub>end</sub>!) * 2<sup>k</sup></code> \r\nwhere <code>len[1], len[2], ..., len[m]</code> are the lengths of each segment of non-infected children that have an infected child at both endpoints, <code>len<sub>start</sub></code> and <code>len<sub>end</sub></code> denote the number of non-infected children with infected child at one endpoint, <code>S</code> is the total length of all segments of non-infected children, and <code>k = (len[1] - 1) + (len[2] - 1) + ... + (len[m] - 1)</code>."
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<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 add_user(struct hash_entry 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