281. Zigzag Iterator
Problem:
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].
Solutions:
public class ZigzagIterator {
Queue<Iterator<Integer>> its = new LinkedList<Iterator<Integer>>();
public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
if (v1.size() > 0) {
its.add(v1.iterator());
}
if (v2.size() > 0) {
its.add(v2.iterator());
}
}
public int next() {
Iterator<Integer> it = its.poll();
int res = it.next();
if (it.hasNext()) {
its.add(it);
}
return res;
}
public boolean hasNext() {
return its.size() > 0;
}
}