Problem 342: Power of Four

https://leetcode.com/problems/power-of-four/?tab=Description

思路

• (num >= 0) && ((num & (num - 1)) == 0) 这两句判断的是“是否是平方”

• $(n)^4 - 1 = (n - 1) (n + 1) (n^2 + 1)$

public class Solution {
    public boolean isPowerOfFour(int num) {
        return (num >= 0) && ((num & (num - 1)) == 0) && ((num - 1) % 3 == 0);
    }
}

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

        return num == 1;
    }
}