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
*/
原文：https://www.cnblogs.com/FLAGyuri/p/12078557.html