计算一个整数二进制中1的个数

这是来自牛客网关于一个二进制的运算

我的思路为每次和1 2 4 .....进行按位与运算就可得到二进制中1的个数

代码如下

/*
 * *new coder ponint to offer 11
 * find bit 1 num of a nunber 
 */

#include <stdio.h>
#include <stdlib.h>

int find1num(int number)
{
    int i = 1,count = 0;

    //while( number / i != 0 ) //if number is negtive we will count error
    
    while( i != 0 )
    {
        if( (number & i) != 0)
        {
            count++;
        }    

        i <<= 1; //i = i * 2
    }
    
    return count;
}

int main()
{

    int number;

    while(1)
    {
        scanf("%d",&number);
        
        int count = find1num(number);

        printf("num of bit 1 is : %d
",count);
    }
}

下面是另外一个大哥的思路，可以明显的减少循环的次数。思路为每次将整数和整数-1的数做与运算这样每次将整数最右边的1变成0，这样做的次数就是整数中含有1的个数，代码如下：

/*
 * *new coder ponint to offer 11
 * find bit 1 num of a nunber
 * time best 
 */

#include <stdio.h>
#include <stdlib.h>

int find1num(int number)
{
    int count = 0;

    
    while( number != 0 )
    {
        count++;
        number  = number & (number - 1); //every time do this will make the rightest 1 of number to 0 so we can do this opration the count 1 of number times
    }
    
    return count;
}

int main()
{

    int number;

    while(1)
    {
        scanf("%d",&number);
        
        int count = find1num(number);

        printf("num of bit 1 is : %d
",count);
    }
}