# Question

Formatted question description: https://leetcode.ca/all/342.html

 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?



# Algorithm

### Diveded by 4

The most direct way is to keep dividing by 4 to see if the final result is 1.

### Mast

First judge whether it is 2 to the Nth power: (num&(num-1)) == 0

Remove the number that is not 4 to the power of N

0x55555555 is to get rid of those power of 2 but not power of 4

so that the single 1 bit always appears at the odd position

The highest 1 of the power of 4 is the counting digit, then we only need to compare it with the previous number (0x55555555) <==> 1010101010101010101010101010101. If the number obtained is still itself, we can be sure that it is a power of 4

### Math

After determining that it is a power of 2, I found that as long as it is a power of 4, it can be divisible by 3 after subtracting 1.

# Code

Java

• 
public class Power_of_Four {

class Solution_3 {
public boolean isPowerOfFour(int num) {
return num > 0 && !(num & (num - 1)) && (num - 1) % 3 == 0;
}
}

class Solution_intuitive {
public boolean isPowerOfFour(int num) {
while (num && (num % 4 == 0)) {
num /= 4;
}
return num == 1;
}
}

class Solution {
public boolean isPowerOfFour(int num) {

return num > 0 && (num&(num-1)) == 0 && (num & 0x55555555) != 0;

}
}
}

• // OJ: https://leetcode.com/problems/power-of-four/
// Time: O(1)
// Space: O(1)
class Solution {
public:
bool isPowerOfFour(int num) {
return __builtin_popcount(num) == 1 && ((~0x55555555 & num) == 0);
}
};

• class Solution(object):
def isPowerOfFour(self, num):
"""
:type num: int
:rtype: bool
"""
return num & (num - 1) == 0 and (num - 1) % 3 == 0