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 <= ) return false;
return (n & (n - )) == ? 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?
public class Solution {
public boolean isPowerOfThree(int n) {
if (n == ) return false;
if (n == ) return true;
if (n > )
return n % == && isPowerOfThree(n / );
else
return false;
}
}
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.
private boolean smartBit(long num){
return (num > ) && ((num & (num - )) == ) && ((num & 0x5555555555555555l) == num);
}
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.