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

Example:

Input:

v1 = [1,2]
v2 = [3,4,5,6] 


Output:
[1,3,2,4,5,6]


Explanation:
 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 givenk1d vectors? How well can your code be extended to such cases?

Clarificationfor the follow up question:
The "Zigzag" order is not clearly defined and is ambiguous fork > 2cases. If "Zigzag" does not look right to you, replace "Zigzag" with "Cyclic". For example:

Input:

[1,2,3]
[4,5,6,7]
[8,9]


Output: 
[1,4,8,2,5,9,3,6,7]
.

public class ZigzagIterator {

    List<Integer> v1;
    List<Integer> v2;
    int index1;
    int index2;
    boolean useV2;
    public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
        this.v1 = v1;
        this.v2 = v2;
        this.index1 = 0;
        this.index2 = 0;
        useV2 = false;
    }

    public int next() {
        if (!useV2 && index1 < v1.size()){
            useV2 = true;
            return v1.get(index1++);
        }
        if (useV2 && index2 < v2.size()){
            useV2 = false;
            return v2.get(index2++);
        }

        return index1 < v1.size() ? v1.get(index1++) : v2.get(index2++);
    }

    public boolean hasNext() {
        return index1 < v1.size() || index2 < v2.size();
    }
}

/**
 * Your ZigzagIterator object will be instantiated and called as such:
 * ZigzagIterator i = new ZigzagIterator(v1, v2);
 * while (i.hasNext()) v[f()] = i.next();
 */