leetcode 231: Power of Two

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Power of Two

Given an integer, write a function to determine if it is a power of two.

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

[思路]

如果是power of two, 则2进制表达中,有且仅有一个1.  可以通过移位来数1的个数, 这里用了一个巧妙的办法, 即判断   N & (N-1) 是否为0.

[CODE]

public class Solution {
    public boolean isPowerOfTwo(int n) {
       return n > 0 && ((n & (n - 1)) == 0 );
    }
}

leetcode 231: Power of Two

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原文地址：http://blog.csdn.net/xudli/article/details/46784163