求二进制数中1的个数

问题：对于一个字节（8bit）的无符号整形变量，求二进制表示中“1”的个数，要求算法的执行效率尽可能高。

#include <iostream>
using namespace std;
//最简单的思路，除2有余数
int fun1(unsigned int a)
{
    int count;
    count = 0;
    while (a)
    {
        if (a % 2)
            count++;
        a /= 2;
    }
    return count;
}
//用位操作更快，但是时间复杂度仍为二进制的长度
int fun2(unsigned int a)
{
    int count;
    count = 0;
    while (a)
    {
        //if (a & 1)
            //count++;
        count += a & 1;
        a = a>>1;
    }
    return count;
}
//O(2)直接给答案，最快，但是空间复杂度会比较高
int fun3(unsigned int a)
{
    int count;
    count = 0;
    int result[20] = {0, 1, 1, 2,
                      1, 2, 2, 3,
                      1, 2, 2, 3,
                      2, 3, 3, 4};
    while (a)
    {
        count += result[a & 15];
        a = a >> 4;
    }
    return count;
}
//这方法想不起来了，时间复杂度跟二进制中1的个数有关，比前二种方法高效，相比第三种方法，没有用到多余的空间
int fun4(unsigned int a)
{
    int count;
    count = 0;
    while (a)
    {
        a &= (a - 1);
        count ++;
    }
    return count;
}
int main()
{
    unsigned int a, b;
    a = 11;
    b = 63;
    //测试
    cout<<fun1(a)<<endl;
    cout<<fun2(a)<<endl;
    cout<<fun3(a)<<endl;
    cout<<fun4(a)<<endl;

    cout<<fun1(b)<<endl;
    cout<<fun2(b)<<endl;
    cout<<fun3(b)<<endl;
    cout<<fun4(b)<<endl;
    system("pause");
    return 0;
}