You could update fibb[n] before returning fib(n-1) + fib(n-2)
int fib(int n)
{
if ( n <= 1 )
{
fibb[1]=n;
return n;
}
if (fibb[n] ! = 0 ) {
//return fibs(n-1)+fibs(n-2);
}
On Sunday, October 21, 2012 7:46:43 AM UTC-4, rahul sharma wrote:
>
> #include<stdio.h>
> int fib(int n)
> {
> if ( n <= 1 )
> return n;
> return fib(n-1) + fib(n-2);
> }
>
> How can we reduce no of computations with the above code....(iterative
> solution not allowed).
>
--
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To view this discussion on the web visit
https://groups.google.com/d/msg/algogeeks/-/AgmaPZ-P3DIJ.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
