leetcode-problemset/leetcode-cn/originData/w3tCBm.json

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        "question": {
            "questionId": "1000230",
            "questionFrontendId": "LCR 003",
            "categoryTitle": "Algorithms",
            "boundTopicId": 910224,
            "title": "比特位计数",
            "titleSlug": "w3tCBm",
            "content": "<p>English description is not available for the problem. Please switch to Chinese.</p>\n",
            "translatedTitle": "比特位计数",
            "translatedContent": "<p>给定一个非负整数 <code>n</code><b>&nbsp;</b>，请计算 <code>0</code> 到 <code>n</code> 之间的每个数字的二进制表示中 1 的个数，并输出一个数组。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1:</strong></p>\n\n<pre>\n<strong>输入: </strong>n =<strong> </strong>2\n<strong>输出: </strong>[0,1,1]\n<strong>解释: \n</strong>0 --&gt; 0\n1 --&gt; 1\n2 --&gt; 10\n</pre>\n\n<p><strong>示例&nbsp;2:</strong></p>\n\n<pre>\n<strong>输入: </strong>n =<strong> </strong>5\n<strong>输出: </strong><code>[0,1,1,2,1,2]\n</code><span style=\"white-space: pre-wrap;\"><strong>解释:</strong>\n</span>0 --&gt; 0\n1 --&gt; 1\n2 --&gt; 10\n3 --&gt; 11\n4 --&gt; 100\n5 --&gt; 101\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>说明 :</strong></p>\n\n<ul>\n\t<li><code>0 &lt;= n &lt;= 10<sup>5</sup></code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>进阶:</strong></p>\n\n<ul>\n\t<li>给出时间复杂度为&nbsp;<code>O(n*sizeof(integer))</code><strong>&nbsp;</strong>的解答非常容易。但你可以在线性时间&nbsp;<code>O(n)</code><strong>&nbsp;</strong>内用一趟扫描做到吗？</li>\n\t<li>要求算法的空间复杂度为&nbsp;<code>O(n)</code>&nbsp;。</li>\n\t<li>你能进一步完善解法吗？要求在C++或任何其他语言中不使用任何内置函数（如 C++ 中的&nbsp;<code>__builtin_popcount</code><strong>&nbsp;</strong>）来执行此操作。</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><meta charset=\"UTF-8\" />注意：本题与主站 338&nbsp;题相同：<a href=\"https://leetcode-cn.com/problems/counting-bits/\">https://leetcode-cn.com/problems/counting-bits/</a></p>\n",
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