update

leetcode-cn/originData/counting-bits.json

@@ -12,7 +12,7 @@
             "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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             "difficulty": "Easy",
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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.",