Check Power of 2

Using O(1) time to check whether an integer n is a power of 2.

Example

For n=4, return true;

For n=5, return false;

奇数总是有超过2个1.

class Solution {
          public boolean checkPowerOf2(int n) {        if(n <= 0) return false;        return (n & (n - 1)) == 0 ? true : false;    }};

 Check Power of 3

Given an integer, write a function to determine if it is a power of three.

Follow up:

Could you do it without using any loop / recursion?

1 public class Solution { 2     public boolean isPowerOfThree(int n) { 3         if (n == 0)    return false; 4  5         if (n == 1) return true; 6  7         if (n > 1) 8             return n % 3 == 0 && isPowerOfThree(n / 3); 9         else10             return false;11     }12 }

Check Power of 4

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.

1 private boolean smartBit(long num){2     return (num > 0) && ((num & (num - 1)) == 0) && ((num & 0x5555555555555555l) == num);3 }

 Explanation:

The magic numbers might seem very confusing. But they are not random (obviously!). Let us start with 0X55555555. Binary representation of 5 is 0101. So 0X55555555 has 16 ones, 16 zeros and the ones,zeros take alternate positions.