CareerCup

zhaoyangster that looks correct.
I have one variation/minor improvement to suggest, use c=c&c-1 instead of c=c>>1 and you only loop once for each bit, instead of all the bit positions.
{{
int bit_swaps_required( int a, int b ) {
unsigned int count = 0;
for( int c = a ^ b; c != 0; c = c & c-1 ) {
++count;
}
return count;
}
}}

saw the solution before: Use XOR

int bit_swaps_required( int a, int b ) {
unsigned int count = 0;
for( int c = a ^ b; c != 0; c = c >> 1 ) {
count += c & 1;
}
return count;
}

Developer your code is working perfectly.Can u explain the logic behind c=c&c-1?

int num_bits(int x,int y)
{
int a,b;
a=x^y;
b=count_ones(a);
return(b);
}
int count_ones(int x)
{
int count;
while((n&(n-1))!=0)
count++;
return(count);
}

int bitswap(int a,int b)
{
    int c = a^b;
    int count = 0;
    while(c!=0)
    {
       if(c&1 ==1)
       count++;
       
       c = c>>1;
    } 
    return count;
}

int num_bits(int x, int y)
{
int count = 0;
while(x != 0 || y != 0)
{
if((x&1)!=(y&1))
{
count++;
}
x = x>>1;
y = y>>1;
}
return count;
}

int bit_flip(int x, int y)
{
int count=0;
z=x^y;
while(z)
{
z&=z-1;
count++;
}
return count;
}

public int flipBits(int a, int b){
        int c = a ^ b;
        int count =0;
	while(x!=0){
		if((x&1)==1){
			count++;
		}
		x= (x>>>1);
	}
        return count;
    }

it goes like:
1) XOR 2 numbers
2) count the number of 1's

I didn't check it for signed integers. Please correct me if am wrong somewhere.