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Leetcode 281. Zigzag Iterator

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Question:

Given two 1d vectors, implement an iterator to return their elements alternately.

For example, given two 1d vectors:

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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].

Explanation: 如果只有两个vector，挺容易的。就用两个指针，然后每次都判断下就行。

Code

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public class ZigzagIterator {
    List<Integer> V1 = new ArrayList<>();
    List<Integer> V2 = new ArrayList<>();
    int index1 = 0;
    int index2 = 0;
    public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
        V1 = v1;
        V2 = v2;
    }
    public int next() {
        if (index1 <= index2) {
            if (index1 < V1.size()) {
                return V1.get(index1++);
            } else {
                return V2.get(index2++);
            }
        } else {
            if (index2 < V2.size()) {
                return V2.get(index2++);
            } else {
                return V1.get(index1++);
            }
        }
    }
    public boolean hasNext() {
        if (index1 < V1.size() || index2 < V2.size()) return true;
        return false;
    }
}

FollowUp: 如果是K个vector，每次判断指针就非常不合理了。这里有一种巧妙的办法，用Queue，或者list，道理相通。 基本思路：把每个list变成iterator，然后放进Queue中，每次poll一个出来，然后得到iterator到的值，如果iterator里面还有值，也就是vector还没有空的话，再排到队伍的最后。不断循环。 巧妙。

Code：

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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 poll = list.remove();
        int result = (Integer)poll.next();
        if(poll.hasNext()) list.add(poll);
        return result;
    }
    public boolean hasNext() {
        return !list.isEmpty();
    }
}