@Ankuj: Yeah, but he asked for it to be recursive. Yours is iterative. Dave
On Aug 9, 9:56 am, Ankuj Gupta <[email protected]> wrote: > we can do it in logn by using binary search approach found > n is the number whose square root has to be > > if(n==1) > return 1; > if(n==0) > return 0; > int low=0,high=n/2,mid,temp; > while(1) > { > mid = (low+high)/2; > temp = mid*mid; > if(temp==n) > return 1; > else if(temp <n) > low = mid+1; > else > high = mid-1; > if(low == mid || high == mid) > return 0; > } > at the end high will have the required square root if not perfect > square or if perfect square mid will have the required answer -- 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.
