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;

        }
    }
}

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