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?

判斷一個數是否是4的冪，如果一個數是4的冪，則這個數的二進制的1在偶數位上。所以先判斷一個數是不是2的冪。然後判斷這個數與0x55555555按位與的結果，0x55555555使用十六位表示的數，它的奇數位上全是1，偶數位上全是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

相關題目：Power of Two

相關題目：Power of Three

[LeetCode] Power of Four