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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].
1 public class ZigzagIterator { 2 3 List<Iterator<Integer>> iters = new ArrayList<>(); 4 int count = 0; 5 6 public ZigzagIterator(List<List<Integer>> lists) { 7 for (List<Integer> v : lists) { 8 if (!v.isEmpty()) iters.add(v.iterator()); 9 } 10 } 11 12 public int next() { 13 int x = iters.get(count).next(); 14 if (!iters.get(count).hasNext()) 15 iters.remove(count); 16 else 17 count++; 18 19 if (iters.size() != 0) 20 count %= iters.size(); 21 return x; 22 } 23 24 public boolean hasNext() { 25 return !iters.isEmpty(); 26 } 27 } 28 29 /** 30 * Your ZigzagIterator object will be instantiated and called as such: 31 * ZigzagIterator i = new ZigzagIterator(v1, v2); while (i.hasNext()) v[f()] = 32 * i.next(); 33 */
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原文地址：http://www.cnblogs.com/beiyeqingteng/p/5743239.html
