leetcode-problemset/leetcode-cn/originData/counting-bits.json

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        "question": {
            "questionId": "338",
            "questionFrontendId": "338",
            "categoryTitle": "Algorithms",
            "boundTopicId": 1431,
            "title": "Counting Bits",
            "titleSlug": "counting-bits",
            "content": "<p>Given an integer <code>n</code>, return <em>an array </em><code>ans</code><em> of length </em><code>n + 1</code><em> such that for each </em><code>i</code><em> </em>(<code>0 &lt;= i &lt;= n</code>)<em>, </em><code>ans[i]</code><em> is the <strong>number of </strong></em><code>1</code><em><strong>&#39;s</strong> in the binary representation of </em><code>i</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 2\n<strong>Output:</strong> [0,1,1]\n<strong>Explanation:</strong>\n0 --&gt; 0\n1 --&gt; 1\n2 --&gt; 10\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> n = 5\n<strong>Output:</strong> [0,1,1,2,1,2]\n<strong>Explanation:</strong>\n0 --&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<p><strong>Constraints:</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<p><strong>Follow up:</strong></p>\n\n<ul>\n\t<li>It is very easy to come up with a solution with a runtime of <code>O(n log n)</code>. Can you do it in linear time <code>O(n)</code> and possibly in a single pass?</li>\n\t<li>Can you do it without using any built-in function (i.e., like <code>__builtin_popcount</code> in C++)?</li>\n</ul>\n",
            "translatedTitle": "比特位计数",
            "translatedContent": "<p>给你一个整数 <code>n</code> ，对于&nbsp;<code>0 &lt;= i &lt;= n</code> 中的每个 <code>i</code> ，计算其二进制表示中 <strong><code>1</code> 的个数</strong> ，返回一个长度为 <code>n + 1</code> 的数组 <code>ans</code> 作为答案。</p>\n\n<p>&nbsp;</p>\n\n<div class=\"original__bRMd\">\n<div>\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 2\n<strong>输出：</strong>[0,1,1]\n<strong>解释：</strong>\n0 --&gt; 0\n1 --&gt; 1\n2 --&gt; 10\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>n = 5\n<strong>输出：</strong>[0,1,1,2,1,2]\n<strong>解释：</strong>\n0 --&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>提示：</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>很容易就能实现时间复杂度为 <code>O(n log n)</code> 的解决方案，你可以在线性时间复杂度 <code>O(n)</code> 内用一趟扫描解决此问题吗？</li>\n\t<li>你能不使用任何内置函数解决此问题吗？（如，C++ 中的&nbsp;<code>__builtin_popcount</code> ）</li>\n</ul>\n</div>\n</div>\n",
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                    "name": "Bit Manipulation",
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                    "code": "class Solution {\npublic:\n    vector<int> countBits(int n) {\n\n    }\n};",
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                    "code": "class Solution(object):\n    def countBits(self, n):\n        \"\"\"\n        :type n: int\n        :rtype: List[int]\n        \"\"\"",
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                    "code": "/**\n * Note: The returned array must be malloced, assume caller calls free().\n */\nint* countBits(int n, int* returnSize) {\n    \n}",
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                    "code": "public class Solution {\n    public int[] CountBits(int n) {\n\n    }\n}",
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                    "code": "/**\n * @param {number} n\n * @return {number[]}\n */\nvar countBits = function(n) {\n\n};",
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                    "lang": "Go",
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                    "lang": "Ruby",
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                    "code": "# @param {Integer} n\n# @return {Integer[]}\ndef count_bits(n)\n\nend",
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                    "lang": "Erlang",
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                    "lang": "Elixir",
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            "hints": [
                "You should make use of what you have produced already.",
                "Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.",
                "Or does the odd/even status of the number help you in calculating the number of 1s?"
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