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].
[分析] 三种实现方案,方案1针对原题,方案2和方案3针对扩展到k个输入链表的情况,其中方案3使用队列很巧妙, 方案2中hasNext()在仅剩一个链表时复杂度是O(k),而方案3始终是O(1)
public class ZigzagIterator { private int active = 0; private List<Iterator<Integer>> iters; public ZigzagIterator(List<Integer> v1, List<Integer> v2) { iters = new ArrayList<Iterator<Integer>>(); iters.add(v1.iterator()); iters.add(v2.iterator()); } public int next() { active = 1 - active; return iters.get(1 - active).next(); } public boolean hasNext() { if (iters.get(active).hasNext()) { return true; } else if (iters.get(1 - active).hasNext()) { active = 1 - active; return true; } else { return false; } } }
// Method 2: 扩展至k,使用数组保存各链表的iterator public class ZigzagIterator { private int total = 2; private int active = 0; private List<Iterator<Integer>> iters; public ZigzagIterator(List<Integer> v1, List<Integer> v2) { iters = new ArrayList<Iterator<Integer>>(); iters.add(v1.iterator()); iters.add(v2.iterator()); } public int next() { int curr = active; active++; if (active == total) active = 0; return iters.get(curr).next(); } public boolean hasNext() { if (iters.get(active).hasNext()) { return true; } else { for (int i = 1; i <= total; i++) { int next = (active + i) % total; if (iters.get(next).hasNext()) { active = next; return true; } } return false; } } }
// Method 3 public class ZigzagIterator { private LinkedList<Iterator<Integer>> queue = new LinkedList<Iterator<Integer>>(); public ZigzagIterator(List<Integer> v1, List<Integer> v2) { if (v1.size() > 0) queue.offer(v1.iterator()); if (v2.size() > 0) queue.offer(v2.iterator()); } public int next() { Iterator<Integer> active = queue.poll(); Integer result = active.next(); if (active.hasNext()) queue.offer(active); return result; } public boolean hasNext() { if (!queue.isEmpty()) return true; else return false; } }