Count number of bits set to 1 in an integer

public class CountBits {

    public static void main(String[] argv){
        CountBits cb = new CountBits();
        for(int i = 0; i < 100 ; i++){
            System.out.println(cb.countBits(i) +" " + cb.countBits2(i));
        }
    }

    /*
        Problem:- Count number of bits set to 1 in an integer
        Solution:- Basic idea is performing x&1 gives you value of the LSB then performing >>> on it
          will push the last bit out and so on till x becomes 0. Time complexity is number of bits in the
         integer
     */
    public int countBits(int x){
       int numberOfBits = 0;
       while (x !=0){
           numberOfBits = numberOfBits + ( x&1);
           x = x>>>1;
       }
       return numberOfBits;
    }

    /*
        Solution: This is little more efficient method in which we just remove the last bit that is
        set to 1, that way the time complexity is equal to number of bits set to 1
     */
    public int countBits2(int x){
        int numberOfBits = 0;
        while (x !=0){
           x =  x& (x-1);
            numberOfBits ++;
        }
        return numberOfBits;
    }
}