# 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