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_281.java
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_281.java
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package com.fishercoder.solutions;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
/**
* 281. Zigzag Iterator
*
* 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].
*/
public class _281 {
public static class Solution1 {
public static class ZigzagIterator {
private Iterator<Integer> i;
private Iterator<Integer> j;
private Iterator<Integer> tmp;
public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
i = v2.iterator();
j = v1.iterator();
}
public int next() {
if (j.hasNext()) {
tmp = j;
j = i;
i = tmp;
}
return i.next();
}
public boolean hasNext() {
return i.hasNext() || j.hasNext();
}
}
}
public static class Solution2 {
public static class ZigzagIterator {
Queue<Iterator<Integer>> queue;
public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
queue = new LinkedList<>();
if (v1 != null && !v1.isEmpty()) {
Iterator<Integer> iterator1 = v1.iterator();
queue.offer(iterator1);
}
if (v2 != null && !v2.isEmpty()) {
Iterator<Integer> iterator2 = v2.iterator();
queue.offer(iterator2);
}
}
public boolean hasNext() {
return !queue.isEmpty();
}
public int next() {
Iterator<Integer> iterator = queue.poll();
int next = iterator.next();
if (iterator.hasNext()) {
queue.offer(iterator);
}
return next;
}
}
}
}