That's same as what I said... In first scan, find the length and in second scan, find the required element. This seems to be only tortoise-tortoise scan.
The other answer (your and sanjay's answer) is more efficient with worst case operations half of what is required in first answer. complexity will be o(n) in both cases. (my bad. i had to edit. i mentioned complexity o(n^2) incorrectly.)
Thanks for the problem.