Counting Number of set bits in a number

1. Method 1

Complexity= O(n)

int numOfOnes(int n){
  int count=0;
  while(n!=0){
   if((n & 1) == 1)
    count++;
   n=n>>>1;
  }
  return count;   
 }

Method 2:
The method works as follows:

The operation:
n=n&(n-1)
resets the last set bit of n.

So the number of time the above operation can be done before n is equal to 0 are the number of set bits in n originally.

Complexity= O(m) where m is the number of set bits in n.

int numOfOnes2(int n){
 int count=0;
 while(n!=0){
  n=n&(n-1);   
  count++;
 }
 return count;
}

Also note that if count above is 1, then number n is a power of 2.