题解 | #二进制中1的个数#

//方法1
//int NumberOf1(unsigned int n ) 
//{
//    int count = 0;
//    while(n)
//    {
//        if ( 1==n%2)
//        {
//            count++;
//        }
//        n=n/2;
//    }
//    return count;
//}
//方法2
//int NumberOf1(unsigned int n )
//{
//    int count = 0;
//    for(int i = 0;i<32;i++)//一共32位，循环32次
//    {
//        if((n&1)==1)
//00000000000000000000000000000001----1的补码
//00000000000000000000000000001010----10的补码
//00000000000000000000000000000000----10&1==0
//00000000000000000000000000000101----10>>1
//00000000000000000000000000000001----(10>>1)&1==1   count+1
//测一位右移有一位
//        {
//            count++;
//        }
//        n = n>>1;
//    }
//    return count;
//}
//方法3
int NumberOf1(unsigned int n )
{
    int count = 0;
    while(n)
    {
//00000000000000000000000000001010----10的补码----n=10
//00000000000000000000000000001001----(10-1)的补码
//00000000000000000000000000001000----10&(10-1)==8---n=8
//00000000000000000000000000000111----(8-1)
//00000000000000000000000000000000----0
        n=n&(n-1);
        count++;
    }
    return count;
}