判断一个数是否是2的整数次幂

public class IsPowerOf2 {
    public static void main(String[] args) {
        //判断一个数是否为2的整数次幂
        IsPowerOf2 isPowerOf2 = new IsPowerOf2();
        System.out.println(isPowerOf2.isPowerOf2(64));
        System.out.println(isPowerOf2.isPowerOf2ForMethodOne(64));
        System.out.println(isPowerOf2.isPowerOf2ForMethodTwo(64));
    }

    //普通写法，时间复杂度log(n)
    private boolean isPowerOf2(int num) {
        int temp = 1;
        while (temp <= num) {
            if (temp == num) {
                return true;
            }
            temp = temp * 2;
        }
        return false;
    }

    
    private int[] transToByte(int num) {
        int[] bytes = new int[32];
        for (int i = 0; i < bytes.length; i++) {
            if ((num & (int) (Math.pow(2, i))) != 0) {
                bytes[i] = 1;
            }
        }
        return bytes;
    }

    //先转成二进制数，再判断是否只包含一个1
    private boolean isPowerOf2ForMethodOne(int num) {
        int[] binary = transToByte(num);
        if (num != 1 && binary[0] == 1) {
            return false;
        }
        int count = 0;
        for (int i = 0; i < binary.length; i++) {
            if (binary[i] == 1) {
                count++;
            }
        }
        if (count == 2) {
            return true;
        } else {
            return false;
        }
    }

    //位运算，时间复杂度为O(1)。牛逼啊
    private boolean isPowerOf2ForMethodTwo(int num) {
        return (num & num - 1) == 0;
    }

}