Give a very good method to count the number of ones in a "n" (e.g. 32) bit number.

Give a very good method to count the number of ones in a "n" (e.g. 32) bit number.

 

 

 

 

 

 

 

public class Count {

 public static void main(String args[])
 {
  int count=0;
     int num=1101011011;
    

  for(int i=0;i<6;i++)//统计的位数
  {
   if(num%Math.pow(10,i+1)==Math.pow(10,i))
    count++;  
   num=(int) (num-num%Math.pow(10,i+1));
   
  }
  
  System.out.print(count);
  
 }
 
}