Given an integer (signed 32 bits), write a function to check whether it is a power of 4.
Example:
Given num = 16, return true. Given num = 5, return false.
Follow up: Could you solve it without loops/recursion?
判断一个数是否是4的幂,如果一个数是4的幂,则这个数的二进制的1在偶数位上。所以先判断一个数是不是2的幂。然后判断这个数与0x55555555按位与的结果,0x55555555使用十六位表示的数,它的奇数位上全是1,偶数位上全是0。
class Solution { public: bool isPowerOfFour(int num) { if (num <= 0) return false; if (!(num & (num - 1))) if (num & 0x55555555) return true; return false; } }; // 3 ms
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