Your Second approach is cool :) On Wed, Dec 22, 2010 at 1:06 PM, juver++ <[email protected]> wrote:
> Use bits manipulation tricks. > 1. There is a way to remove a group of consecutive 1's from the right: A = > n & (n + 1). Then check if A==0 then OK. > 2. Second approach: B=n+1, check if B & (B-1) (this checks if B is a power > of 2, so it contains only 1 set bit) is zero then OK. > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]<algogeeks%[email protected]> > . > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- B. Praveen -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
