【LeetCode 0231】【位运算】2的N次方
- Power of Two
Given an integer n, return true if it is a power of two. Otherwise, return false.
An integer n is a power of two, if there exists an integer x such that n == 2^x.
Example 1:
**Input:** n = 1
**Output:** true
**Explanation: **2^0 = 1
Example 2:
**Input:** n = 16
**Output:** true
**Explanation: **2^4 = 16
Example 3:
**Input:** n = 3
**Output:** false
Constraints:
-2^31 <= n <= 2^31 - 1
Follow up: Could you solve it without loops/recursion?
Idea
首先 2^n 不可能是0或者负数
其次 2^n 作为正数，其二进制位有且只有一个位是1
JavaScript Solution
/**
* @param {number} n
* @return {boolean}
*/
var isPowerOfTwo = function(n) {
return !(n<=0) && !(n&(n-1))
};
原文地址：https://blog.csdn.net/avenccssddnn/article/details/140579713
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