Number of 1 Bits

Write a function that takes an unsigned integer and returns the number of ’1’ bits it has (also known as the Hamming weight).

For example, the 32-bit integer ’11’ has binary representation 00000000000000000000000000001011, so the function should return 3.
题目的意思实际上就是找出一个整数（无论是正数还是负数）的二进制码中1的个数。

思路：如果是正数的话，就直接计算出1的个数即可，但是如果是负数的话，就先把它按位取反得到其反码，然后求出反码的1的个数，很显然，如果是负数的二进制码中1的个数就是32减去反码中1的个数。

public class Solution {    // you need to treat n as an unsigned value    public int hammingWeight(int n) {            int sign=0;        if(n<0)             {               sign=1;              n=~n;            }        int r = n % 2;        int b = n / 2;        int count = 0;        while (b != 0) {            if (r == 1)                count++;            r = b % 2;            b = b / 2;        }        if (r == 1)            count++;        if(sign==1) count=32-count;        return count;    }}