0231._power_of_two.md

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• Solution 1 位运算，按位与

Links:

1. https://leetcode.com/problems/power-of-two/
2. https://leetcode-cn.com/problems/power-of-two/

Solution 1 位运算，按位与

2^x	n	n - 1	n & (n - 1)
2^0	0001	0000	(0001) & (0000) == 0
2^1	0010	0001	(0010) & (0001) == 0
2^2	0100	0011	(0100) & (0011) == 0
2^3	1000	0111	(1000) & (0111) == 0

class Solution:
    def isPowerOfTwo(self, n):
        return n > 0 and n & (n - 1) == 0