求二进制中1的个数

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;

class Solution {
public:
    int NumberOf1(int n) {
        unsigned sn = n;
　　　　　//常规做法
        int ret = 0;
        while (sn) {
            ret++;
            sn = (sn - 1) & sn;
        }

        sn = n;
        unsigned tmp;
        tmp = (0x55555555) & sn;
        sn = tmp + ((sn & 0xaaaaaaaa) >> 1);
        //printf("%x\n", sn);

        tmp = (0x33333333) & sn;
        sn = tmp + ((sn & 0xCCCCCCCC) >> 2);
        //printf("%x\n", sn);

        tmp = (0x0F0F0F0F) & sn;
        sn = tmp + ((sn & 0xF0F0F0F0) >> 4);
        //printf("%x\n", sn);

        tmp = (0x00FF00FF) & sn;
        sn = tmp + ((sn & 0xFF00FF00) >> 8);
        //printf("%x\n", sn);

        tmp = (0x0000FFFF) & sn;
        sn = tmp + ((sn & 0xFFFF0000) >> 16);
        //printf("%x\n", sn);

        return sn;
    }
};

int main() {
    Solution so;
    int n;
    cin >> n;
    cout << so.NumberOf1(n) << endl;
    return 0;
}