leetcode 281: Zigzag Iterator

Zigzag Iterator

Total Accepted: 964 Total Submissions: 2714 Difficulty: Medium

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 bynext 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].

[思路]

iterator都放到一个list里, 用一个count循环,

[CODE]

public class ZigzagIterator {    List<Iterator<Integer> > iters = new ArrayList<Iterator<Integer> >();         int count = 0;    public ZigzagIterator(List<Integer> v1, List<Integer> v2) {        if( !v1.isEmpty() ) iters.add(v1.iterator());        if( !v2.isEmpty() ) iters.add(v2.iterator());    }    public int next() {        int x = iters.get(count).next();        if(!iters.get(count).hasNext()) iters.remove(count);        else count++;                if(iters.size()!=0) count %= iters.size();        return x;    }    public boolean hasNext() {        return !iters.isEmpty();    }}/** * Your ZigzagIterator object will be instantiated and called as such: * ZigzagIterator i = new ZigzagIterator(v1, v2); * while (i.hasNext()) v[f()] = i.next(); */