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51 lines
1.5 KiB
HTML
51 lines
1.5 KiB
HTML
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<p>给定一个非负整数 <code>n</code><b> </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>
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<p> </p>
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<p><strong>示例 1:</strong></p>
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<pre>
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<strong>输入: </strong>n =<strong> </strong>2
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<strong>输出: </strong>[0,1,1]
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<strong>解释:
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</strong>0 --> 0
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1 --> 1
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2 --> 10
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</pre>
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<p><strong>示例 2:</strong></p>
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<pre>
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<strong>输入: </strong>n =<strong> </strong>5
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<strong>输出: </strong><code>[0,1,1,2,1,2]
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</code><span style="white-space: pre-wrap;"><strong>解释:</strong>
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</span>0 --> 0
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1 --> 1
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2 --> 10
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3 --> 11
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4 --> 100
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5 --> 101
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</pre>
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<p> </p>
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<p><strong>说明 :</strong></p>
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<ul>
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<li><code>0 <= n <= 10<sup>5</sup></code></li>
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</ul>
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<p> </p>
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<p><strong>进阶:</strong></p>
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<ul>
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<li>给出时间复杂度为 <code>O(n*sizeof(integer))</code><strong> </strong>的解答非常容易。但你可以在线性时间 <code>O(n)</code><strong> </strong>内用一趟扫描做到吗？</li>
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<li>要求算法的空间复杂度为 <code>O(n)</code> 。</li>
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<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 <code>__builtin_popcount</code><strong> </strong>）来执行此操作。</li>
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</ul>
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<p> </p>
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<p><meta charset="UTF-8" />注意：本题与主站 338 题相同：<a href="https://leetcode-cn.com/problems/counting-bits/">https://leetcode-cn.com/problems/counting-bits/</a></p>
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