Description

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

Example

Example1

Input: v1 = [1, 2] and v2 = [3, 4, 5, 6]
Output: [1, 3, 2, 4, 5, 6]
Explanation: 
By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1, 3, 2, 4, 5, 6].

Example2

Input: v1 = [1, 1, 1, 1] and v2 = [3, 4, 5, 6]
Output: [1, 3, 1, 4, 1, 5, 1, 6]

思路：用双指针求解即可，交叉返回两数组当前位置的数，当一个数组被访问完后，就一直访问另一个数组

public class ZigzagIterator {
    
    public Iterator<Integer> it1;
    public Iterator<Integer> it2;
    public int turns;

    /**
     * @param v1 v2 two 1d vectors
     */
    public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
        // initialize your data structure here.
        this.it1 = v1.iterator();
        this.it2 = v2.iterator();
        turns = 0;
    }

    public int next() {
        // Write your code here
        turns++;
        if((turns % 2 == 1 && it1.hasNext()) || (!it2.hasNext())) {
            return it1.next();
        } else if((turns % 2 == 0 && it2.hasNext()) || (!it1.hasNext())) {
            return it2.next();
        }
        return -1;  
    }

    public boolean hasNext() {
        // Write your code here
        return it1.hasNext() || it2.hasNext();        
    }
}


/**
 * Your ZigzagIterator object will be instantiated and called as such:
 * ZigzagIterator solution = new ZigzagIterator(v1, v2);
 * while (solution.hasNext()) result.add(solution.next());
 * Output result
 */