281 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?

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].

k = 2

这道题可以直接调用list的iterator来实现，只要用一个计数来确定应该用哪个iteraor即可。当然也可以自己用两个index来实现，但是iterator在内存中占用空间更小，并且更方便操作

public class ZigzagIterator {
    Iterator<Integer> it1;
    Iterator<Integer> it2;
    int count;
    
    public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
        it1 = v1.iterator();
        it2 = v2.iterator();
        count = 0;
    }
    
    public int next() {
        count++;
        if ((count % 2 == 1 && it1.hasNext()) || !it2.hasNext()) {
            return it1.next();
        } else if ((count % 2 == 0 && it2.hasNext()) || !it1.hasNext()) {
            return it2.next();
        } else {
            return -1;
        }
    }
    
    public boolean hasNext() {
        return it1.hasNext() || it2.hasNext();
    }
}

K > 2

和k = 2类似，只是用一个list来存k个iterator（如果iterator中没有元素则不用加入）。用一个count%list_size来确定应该用哪个iterator。若其中一个iterator里面的数已经被取完（即该iterator的hasNext()为false），则将其从list中移除，并将count设置为count%new list_size（如果new list_size != 0）

public class ZigzagIterator2 {
    List<Iterator<Integer>> vec;
    int count;
    
    public ZigzagIterator2(List<List<Integer>> vecs) {
        vec = new ArrayList<>();
        for (List<Integer> list : vecs) {
            if (list.size() > 0) {
                vec.add(list.iterator());
            }
        }
        count = 0;
    }
    
    public int next() {
        int res = vec.get(count).next();
        if (vec.get(count).hasNext()) {
            count = (count + 1) % vec.size();
        } else {
            vec.remove(count);
            if (vec.size() > 0) {
                count %= vec.size();
            }
        }
        return res;
    }
    
    public boolean hasNext() {
        return vec.size() > 0;
    }
}

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