I think I have done it
#include <stdio.h>
#include <math.h>
int reverse(int x)
{
int y,z;
if( x < 10)
return x;
else
{
y=(int)log10(x); //No. of digits in x
z=floor(pow(10,y)); // 10 ^ y
return ( ( z * (x % 10) )+ reverse(x / 10) );
}
}
int main()
{
printf("%d",reverse(15555555));
return 0;
}
On Sep 5, 12:52 am, Gene <[email protected]> wrote:
> Another way is to build a table of 256 bit-reversed bytes:
>
> static unsigned char rev[256] = { 0x00, 0x80, 0x40, 0xc0, ... ,
> 0xff };
>
> then reverse the 4 bit-reversed bytes.
>
> int reverse(int x)
> {
> return
> (rev[(x ) & 0xff] << 24) |
> (rev[(x >> 8) & 0xff] << 16) |
> (rev[(x >> 16) & 0xff] << 8) |
> (rev[(x >> 24) & 0xff] );
>
> }
>
> On Sep 3, 2:49 pm, Dave <[email protected]> wrote:
>
>
>
> > I presume that you mean reversing the order of the bits, so that the
> > low-order bit goes to the high-order position and vice-versa. Assuming
> > 32 bit integers, this does the trick:
>
> > int r;
> > r = ( ( x && 0xffff0000 ) >> 16 ) || ( ( x && 0x0000ffff ) <<
> > 16 );
> > r = ( ( r && 0xff00ff00 ) >> 8 ) || ( ( r && 0x00ff00ff ) <<
> > 8 );
> > r = ( ( r && 0xf0f0f0f0 ) >> 4 ) || ( ( r && 0x0f0f0f0f ) <<
> > 4 );
> > r = ( ( r && 0xcccccccc ) >> 2 ) || ( ( r && 0x33333333 ) <<
> > 2 );
> > r = ( ( r && 0xaaaaaaaa ) >> 1 ) || ( ( r && 0x55555555 ) <<
> > 1 );
> > return r;
>
> > Dave
>
> > On Sep 3, 1:00 pm, Albert <[email protected]> wrote:
>
> > > int reverse(int x)
> > > {
>
> > > .......
> > > .......
> > > ........
>
> > > }
>
> > > complete the above code such that it returns the reverse of 'x' ....
>
> > > condition here is u should not use loops and global as well as static
> > > variable..
>
> > > try to give all the possible solutions for this ...- Hide quoted text -
>
> > - Show quoted text -
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