[Solved] How would you change this to a while loop and make it have a better run-time complexity?

Question

Yes, you can.
Your algorithm is running in O(n*m) time, and is basically saying:

Are the two lists denoted as A and B have any common items?

This can be done more efficiently if you first store one list into a set, which is implemented with more efficient data structure (Usually self balancing binary tree or hash table).

Then, you simply iterate the second list and check for matches.

Pseudo code:

Algorithm foo(A,n,B,m)
    Input: arrays of integers, A of length n and B of length m
    Output: true or false
    s = new Hash/Tree set
    for i := 0 to n
        s.add(A[i])
    endfor
    for j := 0 to m
        if s.contains(B[j])
           return true
        endif
    endfor
    return false

The new version will run in O(logn*(n+m)) or O(n+m), depending if you used tree based set or hash based set.

Accepted Answer