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

1 /* 2  * *new coder ponint to offer 11 3  * find bit 1 num of a nunber  4  */ 5  6 #include 
     
       7 #include 
      
        8  9 int find1num(int number)10 {11     int i = 1,count = 0;12 13     //while( number / i != 0 ) //if number is negtive we will count error14     15     while( i != 0 )16     {17         if( (number & i) != 0)18         {19             count++;20         }    21 22         i <<= 1; //i = i * 223     }24     25     return count;26 }27 28 int main()29 {30 31     int number;32 33     while(1)34     {35         scanf("%d",&number);36         37         int count = find1num(number);38 39         printf("num of bit 1 is : %d\n",count);40     }41 }
      
     

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：

1 /* 2  * *new coder ponint to offer 11 3  * find bit 1 num of a nunber 4  * time best  5  */ 6  7 #include 
     
       8 #include 
      
        9 10 int find1num(int number)11 {12     int count = 0;13 14     15     while( number != 0 )16     {17         count++;18         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 times19     }20     21     return count;22 }23 24 int main()25 {26 27     int number;28 29     while(1)30     {31         scanf("%d",&number);32         33         int count = find1num(number);34 35         printf("num of bit 1 is : %d\n",count);36     }37 }