count number of bits of '1' from 1 to n

Q:
from 1 to n, count the number of bits that were set to 1.
For example,
n == 1,  return 1,
n == 2, return 3,
n == 3, return 5
n==4, return 6



A:
solution  1:  一个个数，
Solution 2：把n 从2^j 一个个整除，来计算。

 private static int numberOf1Bits(int n) {
        int mask = 0; // set the mask as the left part of nPlusOne
        int nBits = 0; // counts the bits from 1 to n in their binary form

        for (int i = 30; i >= 0; i--) {
            int power2 = 1 << i;
            if(n/power2<1)  // this two lines can be commented
                continue;    // can be commented
            mask |= (n & power2);
            int nRepeatOnes = n / (power2<<1) * power2 ;
            int nLeftOnes = ( n>>i & 1)  * (n - mask+1);

            nBits += nRepeatOnes + nLeftOnes;
        }
        return nBits;
    }

Mistakes:
1:  这里，要除的时候，跟n+1无关，而应该是n / 上一位的power，因此要power2<<1