求二进制中1的个数

给定一个已知的整数n,求该数二进制1的个数

仔细读题，可以发现，其实求该数二进制1的个数，只需依次判断最后一位数是否为1，然后循环右移即可。按照该思路写下如下所示代码：

int count(unsigned char n)
{
    int sum=0;
    while(n)
    {
        sum+=n&0x01;
        n>>=1;
    }
    return sum;
}

还可以按照求余法来实现

int count(unsigned char n)
{
    int sum=0;
    while(n)
    {
        if(n%2==1)
            sum++;
        n/=2;
    }
    return sum;
}

1中所述方法始终是循环移动8位，

int count(unsigned char n)
{
    int sum=0;
    while(n)
    {
        n&=(n-1);
        sum++;
    }
    return sum;
}

int table[256]={
            0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4,               
            1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,                
            1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,               
            2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,                
            1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,                
            2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,                
            2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,                
            3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,               
            1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,                
            2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,                
            2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,                
            3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,               
            2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,                
            3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,                
            3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,                
            4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8, 
            };

int count(unsigned char n)
{
    return table[n];
}