Zigzag Iterator
Total Accepted: 7833 Total Submissions: 18659 Difficulty: Medium
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].
思路
- 考察Iterator
- 利用其本身的iterator
public class ZigzagIterator {
Iterator<Integer>[] its;
int pos;
public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
its = new Iterator[]{v1.iterator(), v2.iterator()};
pos = 0;
}
public int next() {
int next = its[pos].next();
pos = (pos == its.length - 1) ? 0 : pos + 1;
return next;
}
public boolean hasNext() {
if (its[pos].hasNext()) return true;
for (int i = pos + 1; i < its.length; i++) {
if (its[i].hasNext()) {
pos = i;
return true;
}
}
for (int i = 0; i < pos; i++) {
if (its[i].hasNext()) {
pos = i;
return true;
}
}
return false;
}
}