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].

Keep the invariant in mind.

//invariant keep row, col always points to next available element
class ZigzagIterator {
public:
    vector<vector<int>> data;
    int row, col;
    bool isEnd;
    ZigzagIterator(vector<int>& v1, vector<int>& v2) {
        data.push_back(v1);
        data.push_back(v2);
        row = col = 0;
        isEnd = false;
        while(row < data.size() && data[row].size() == 0){
            row++;
        }
        if(row == data.size()){
            isEnd = true;
        }
    }

    int next() {
        int ans = data[row][col];
        row++;
        while(row < data.size() && data[row].size() <= col){
            row++;
        }
        if(row == data.size()){
            row = 0;
            col++;
        }
        while(row < data.size() && data[row].size() <= col){
            row++;
        }
        if(row == data.size()){
            isEnd = true;
        }
        return ans;
    }

    bool hasNext() {
        return !isEnd;
    }
};

/**
 * Your ZigzagIterator object will be instantiated and called as such:
 * ZigzagIterator i(v1, v2);
 * while (i.hasNext()) cout << i.next();
 */