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?

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].

分析

这种题的通用解法就是用一个Queue来装所有iterator, next()的时候，pop一个iterator, 取对应iterator.next(), 如果之后iterator.hasNext()再把该iterator push回到Queue中。

关键就是Queue中所存在的iterator必须hasNext()，所有写Constructor时也要注意。

复杂度

time: O(n), space: O(n)

代码

public class ZigzagIterator {
    Queue<Iterator<Integer>> q;
    
    public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
        q = new LinkedList<Iterator<Integer>>();
        Iterator<Integer> it1 = v1.iterator();
        Iterator<Integer> it2 = v2.iterator();
        
        // 保证Queue中的Iterator hasNext().
        if (it1.hasNext())
            q.add(it1);
        if (it2.hasNext())
            q.add(it2);
    }

    public int next() {
        Iterator<Integer> it = q.remove();
        int res = it.next();
        if (it.hasNext()) {
            q.add(it);
        }
        return res;
    }

    public boolean hasNext() {
        return !q.isEmpty();
    }
}