Leetcode: Zigzag Iterator

Given two 1d vectors, implement an iterator to return their elements alternately.

For example, given two 1d vectors:

v1 = [1, 2]
v2 = [3, 4, 5, 6]
By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1, 3, 2, 4, 5, 6].

Follow up: What if you are given k 1d vectors? How well can your code be extended to such cases?

Clarification for the follow up question - Update (2015-09-18):
The "Zigzag" order is not clearly defined and is ambiguous for k > 2 cases. If "Zigzag" does not look right to you, replace "Zigzag" with "Cyclic". For example, given the following input:

[1,2,3]
[4,5,6,7]
[8,9]
It should return [1,4,8,2,5,9,3,6,7].

 1 public class ZigzagIterator {
 2     List<Integer> l1;
 3     List<Integer> l2;
 4     int cnt1;
 5     int cnt2;
 6     
 7     public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
 8         this.l1 = v1;
 9         this.l2 = v2;
10         this.cnt1 = 0;
11         this.cnt2 = 0;
12     }
13 
14     public int next() {
15         if (cnt1<l1.size() && cnt2<l2.size()) {
16             if (cnt1 == cnt2) {
17                 cnt1++;
18                 return l1.get(cnt1-1);
19             }
20             else {
21                 cnt2++;
22                 return l2.get(cnt2-1);
23             }
24         }
25         else if (cnt1 < l1.size()) {
26             cnt1++;
27             return l1.get(cnt1-1);
28         }
29         else if (cnt2 < l2.size()) {
30             cnt2++;
31             return l2.get(cnt2-1);
32         }
33         else return Integer.MIN_VALUE;
34     }
35 
36     public boolean hasNext() {
37         return (cnt1<l1.size()) || (cnt2<l2.size());
38     }
39 }
40 
41 /**
42  * Your ZigzagIterator object will be instantiated and called as such:
43  * ZigzagIterator i = new ZigzagIterator(v1, v2);
44  * while (i.hasNext()) v[f()] = i.next();
45  */