Recursive function to count adjacent duplicates

Question

I want the recursive function to check the value in the current node equals the value in the next node and increment by 1 else don't increment and move on until the end of the list. So for, a list of 1, 2, 2, 3 , 1, 1. It should return 2 as 2, 2 is one adjacent duplicate and 1, 1 is another adjacent duplicate.

I can't figure out how to handle the false case when the value of the current isn't equal to the value of the next. Basically, not incrementing.

This is my code so far...

int fn(Node l) {
    if (l == null)
        return 0;
    else 
        return (l.value == l.next.value) ? (1 + fn(l.next)) : ;
}

I'm not sure what you are trying to accomplish. Could you please provide a better description of what you are trying to do? — Joel Westberg
– Joel Westberg, Commented Oct 13, 2012 at 18:04
If the value in the current node = value in the next node increment by 1 else don't increment and move on until the end of the list. So for, a list of 1, 2, 2, 3 , 1, 1. It should return 2. 2,2 repeats and 1,1 repeats. — nitiger
– nitiger, Commented Oct 13, 2012 at 18:07

Stuart Wakefield · Accepted Answer · 2012-10-13 18:30:09Z

2

The function needs to be called again in either case, for the false case you would not add 1 to the return value, i.e.

return (l.value == l.next.value) ? (1 + fn(l.next)) : fn(l.next);

You should also check that the l.next is not null first. So you could rewrite the function...

int fn(Node l) {
   if (l == null || l.next == null)
       return 0;
   return (l.value == l.next.value ? 1 : 0) + fn(l.next);
}

edited Oct 13, 2012 at 18:30

Stuart Wakefield

6,40431 silver badges39 bronze badges

answered Oct 13, 2012 at 18:07

perh

1,72811 silver badges14 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

Recursive function to count adjacent duplicates

1 Answer 1

Comments

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Related