342. Power of Four

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?

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

首先，可以用万能法。代码如下：

public class Solution {
    public boolean isPowerOfFour(int num) {
        if (num > 1) {
            while (num % 4 == 0) {
                num ／= 4;
            }
        }
        return num == 1;
    }
}

或者用数学证明法，代码如下：

public class Solution {
    /*
    (4^n - 1) % 3 == 0
    another proof:
    (1) 4^n - 1 = (2^n + 1) * (2^n - 1)
    (2) among any 3 consecutive numbers, there must be one that is a multiple of 3
    among (2^n-1), (2^n), (2^n+1), one of them must be a multiple of 3, and (2^n) cannot be the one, therefore either (2^n-1) or (2^n+1) must be a multiple of 3, and 4^n-1 must be a multiple of 3 as well.
    */
    public boolean isPowerOfFour(int num) {
       return num > 0 && (num & (num - 1)) == 0 && (num - 1) % 3 == 0;
    }
}