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 given k 1d vectors? How well can your code be extended to such cases?
Clarification for the follow up question:
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:
Input:
[1,2,3]
[4,5,6,7]
[8,9]
Output: [1,4,8,2,5,9,3,6,7].
ไปฃ็
Approach #1
publicclassZigzagIterator {LinkedList<Iterator> list;publicZigzagIterator(List<Integer> v1,List<Integer> v2) { list =newLinkedList<Iterator>();if (!v1.isEmpty()) list.add(v1.iterator());if (!v2.isEmpty()) list.add(v2.iterator()); }publicintnext() {Iterator poll =list.remove(); // ๅคดint result = (Integer) poll.next();if (poll.hasNext) list.add(poll);return result; }publicbooleanhasNext() {return!list.isEmpty(); }}/** * Your ZigzagIterator object will be instantiated and called as such: * ZigzagIterator i = new ZigzagIterator(v1, v2); * while (i.hasNext()) v[f()] = i.next(); */