281 Zigzag Iterator

281. Zigzag Iterator

1. Question

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 callingnextrepeatedly untilhasNextreturnsfalse, the order of elements returned bynextshould be:[1, 3, 2, 4, 5, 6].

Follow up: What if you are givenk1d 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 fork > 2cases. 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].

2. Implementation

思路: 用LinkedList或者Queue存每个输入list的iterator，这样就可以解决follow-up里的k 个vector的问题

public class ZigzagIterator {
    LinkedList<Iterator> list;

    public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
        list = new LinkedList<Iterator>();
        if (!v1.isEmpty()) {
            list.add(v1.iterator());
        }

        if (!v2.isEmpty()) {
            list.add(v2.iterator());
        }
    }

    public int next() {
        Iterator curIT = list.remove();
        int res = (Integer)curIT.next();
        if (curIT.hasNext()) {
            list.add(curIT);
        }
        return res;
    }

    public boolean hasNext() {
        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();
 */

3. Time & Space Complexity

时间复杂度: next(): O(1), hasNext(): O(1), 空间复杂度O(k), k是输入list的个数