[LeetCode] 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.
n是正数并且n的二进制只有一个1,那么n就是2的幂。所以如果 n > 0 && n - lowbit(n) == 0,那么就return true。lowbit(n) = n & -n,作用是提取出n的最低位的1来,比如lowbit(111000) = 001000。或者 n > 0 && !(n & (n - 1)),n & (n - 1)的作用是去掉最后面的一位1。
1 class Solution { 2 public: 3 bool isPowerOfTwo(int n) { 4 return n > 0 && (n - (n & -n) == 0); 5 } 6 };