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add leetcode problem-cn part6
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算法题(国内版)/problem (Chinese)/2 的幂 [power-of-two].html
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算法题(国内版)/problem (Chinese)/2 的幂 [power-of-two].html
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<p>给你一个整数 <code>n</code>,请你判断该整数是否是 2 的幂次方。如果是,返回 <code>true</code> ;否则,返回 <code>false</code> 。</p>
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<p>如果存在一个整数 <code>x</code> 使得 <code>n == 2<sup>x</sup></code> ,则认为 <code>n</code> 是 2 的幂次方。</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 = 1
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<strong>输出:</strong>true
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<strong>解释:</strong>2<sup>0</sup> = 1
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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 = 16
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<strong>输出:</strong>true
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<strong>解释:</strong>2<sup>4</sup> = 16
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</pre>
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<p><strong>示例 3:</strong></p>
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<pre>
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<strong>输入:</strong>n = 3
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<strong>输出:</strong>false
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</pre>
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<p><strong>示例 4:</strong></p>
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<pre>
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<strong>输入:</strong>n = 4
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<strong>输出:</strong>true
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</pre>
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<p><strong>示例 5:</strong></p>
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<pre>
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<strong>输入:</strong>n = 5
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<strong>输出:</strong>false
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</pre>
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<p> </p>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>-2<sup>31</sup> <= n <= 2<sup>31</sup> - 1</code></li>
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</ul>
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<p> </p>
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<p><strong>进阶:</strong>你能够不使用循环/递归解决此问题吗?</p>
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