LeetCode – Power of Two (Java)

by

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

Analysis

If a number is power of 2, it’s binary form should be 10…0. So if we right shift a bit of the number and then left shift a bit, the value should be the same when the number >= 10(i.e.,2).

Java Solution 1 

public boolean isPowerOfTwo(int n) {
    if(n<=0) 
        return false;
 
    while(n>2){
        int t = n>>1;
        int c = t<<1;
 
        if(n-c != 0)
            return false;
 
        n = n>>1;
    }
 
    return true;
}

Java Solution 2

If a number is power of 2, then its highly bit is 1 and there is only one 1. Therefore, n & (n-1) is 0.

public boolean isPowerOfTwo(int n) {
    return n>0 && (n&n-1)==0;
}

Java Solution 3

public boolean isPowerOfTwo(int n) {
    return n>0 && n==Math.pow(2, Math.round(Math.log(n)/Math.log(2)));
}

In this solution, the Math.round() method rounds up the number.

4 thoughts on “LeetCode – Power of Two (Java)”

  4. if (n&(n-1)) ==0, means n is power of 2

    public class Solution {
    public boolean isPowerOfTwo(int n) {
    return n>0&& ((n&(n-1))==0);
    }
    }

Leave a Comment