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].
public class ZigzagIterator {
List<Iterator<Integer>> iters = new ArrayList<>();
int count = ;
public ZigzagIterator(List<List<Integer>> lists) {
for (List<Integer> v : lists) {
if (!v.isEmpty()) iters.add(v.iterator());
}
}
public int next() {
int x = iters.get(count).next();
if (!iters.get(count).hasNext())
iters.remove(count);
else
count++;
if (iters.size() != )
count %= iters.size();
return x;
}
public boolean hasNext() {
return !iters.isEmpty();
}
}
/**
* Your ZigzagIterator object will be instantiated and called as such:
* ZigzagIterator i = new ZigzagIterator(v1, v2); while (i.hasNext()) v[f()] =
* i.next();
*/