281. Zigzag Iterator

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].

Solution

public class ZigzagIterator {

     boolean flag;
    Iterator<Integer> iterator1;
    Iterator<Integer> iterator2;

    public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
        iterator1 = v1.iterator();
        iterator2 = v2.iterator();
        flag = true;
    }

    public int next() {
        if (hasNext()) {
            if (!iterator1.hasNext()) {
                return iterator2.next();
            }

            if (!iterator2.hasNext()) {
                return iterator1.next();
            }

            if (flag && iterator1.hasNext()) {
                flag = false;
                return iterator1.next();
            }

            if (!flag && iterator2.hasNext()) {
                flag = true;
                return iterator2.next();
            }
        }
            return Integer.MIN_VALUE;

    }

    public boolean hasNext() {
        return iterator1.hasNext() || iterator2.hasNext();
    }
}

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