LeetCode Zigzag Iterator
原题链接在这里:https://leetcode.com/problems/zigzag-iterator/
题目:
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]
.
题解:
用queue来保存每个list的iterator. next() 是 poll 出que的第一个iterator, return iterator.next(), 若是该iterator还有next, 就再放到queue尾部.
若queue空了,说明都遍历过了,此时hasNext()就返回false.
Time Complexity: next, O(1). hasNext, O(1). Space: O(l), l 是list的个数.
AC Java:
public class ZigzagIterator {
LinkedList<Iterator<Integer>> que;
public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
que = new LinkedList<Iterator<Integer>>();
if(v1.iterator().hasNext()){
que.add(v1.iterator());
}
if(v2.iterator().hasNext()){
que.add(v2.iterator());
}
} public int next() {
Iterator<Integer> it = que.poll();
int cur = it.next();
if(it.hasNext()){
que.add(it);
}
return cur;
} public boolean hasNext() {
return !que.isEmpty();
}
} /**
* Your ZigzagIterator object will be instantiated and called as such:
* ZigzagIterator i = new ZigzagIterator(v1, v2);
* while (i.hasNext()) v[f()] = i.next();
*/
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