Give a very good method to count the number of ones in a 32
bit number.
(caution: looping through testing each bit is not a solution)
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Answer Posted / turk
Here is an O(logn) solution
#include<stdio.h>
int numberOfOnesByte(unsigned char c, int length){
if (c==0)
return 0;
if (c==1)
return 1;
unsigned char left, right;
int lengthP = length/2;
left = c>>lengthP;
right = c-(left<<lengthP);
return numberOfOnesByte(left,length-lengthP)+numberOfOnesByte(right,lengthP);
}
int numberOfOnes(unsigned char *array, int start, int end, int length){
if(length==1){
return numberOfOnesByte(array[start],8);
} else {
int lengthP = length/2;
return numberOfOnes(array,start,start+lengthP-1,lengthP)+
numberOfOnes(array,start+lengthP,end,length-lengthP);
}
}
int main(){
unsigned char array[8] = {0xFF,0XAA,0xFF,0XAA,0xFF,0XAA,0xFF,0XAA};
printf("number of bits %d\n",numberOfOnes(array,0,7,8));
return 0;
}
| Is This Answer Correct ? | 3 Yes | 5 No |
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