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leetcode-cn/originData/prison-cells-after-n-days.json

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             "translatedContent": "<p>8 间牢房排成一排，每间牢房不是有人住就是空着。</p>\n\n<p>每天，无论牢房是被占用或空置，都会根据以下规则进行更改：</p>\n\n<ul>\n\t<li>如果一间牢房的两个相邻的房间都被占用或都是空的，那么该牢房就会被占用。</li>\n\t<li>否则，它就会被空置。</li>\n</ul>\n\n<p>（请注意，由于监狱中的牢房排成一行，所以行中的第一个和最后一个房间无法有两个相邻的房间。）</p>\n\n<p>我们用以下方式描述监狱的当前状态：如果第 <code>i</code> 间牢房被占用，则 <code>cell[i]==1</code>，否则 <code>cell[i]==0</code>。</p>\n\n<p>根据监狱的初始状态，在 <code>N</code> 天后返回监狱的状况（和上述 N 种变化）。</p>\n\n<p>&nbsp;</p>\n\n<ol>\n</ol>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><strong>输入：</strong>cells = [0,1,0,1,1,0,0,1], N = 7\n<strong>输出：</strong>[0,0,1,1,0,0,0,0]\n<strong>解释：\n</strong>下表概述了监狱每天的状况：\nDay 0: [0, 1, 0, 1, 1, 0, 0, 1]\nDay 1: [0, 1, 1, 0, 0, 0, 0, 0]\nDay 2: [0, 0, 0, 0, 1, 1, 1, 0]\nDay 3: [0, 1, 1, 0, 0, 1, 0, 0]\nDay 4: [0, 0, 0, 0, 0, 1, 0, 0]\nDay 5: [0, 1, 1, 1, 0, 1, 0, 0]\nDay 6: [0, 0, 1, 0, 1, 1, 0, 0]\nDay 7: [0, 0, 1, 1, 0, 0, 0, 0]\n\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><strong>输入：</strong>cells = [1,0,0,1,0,0,1,0], N = 1000000000\n<strong>输出：</strong>[0,0,1,1,1,1,1,0]\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ol>\n\t<li><code>cells.length == 8</code></li>\n\t<li><code>cells[i]</code>&nbsp;的值为 <code>0</code> 或 <code>1</code>&nbsp;</li>\n\t<li><code>1 &lt;= N &lt;= 10^9</code></li>\n</ol>\n",
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