标签:
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]
.
链接: http://leetcode.com/problems/zigzag-iterator/
题解:
Zigzag Iterator。最简单的就是用两个index分别遍历两个list,然后用一个boolean变量来控制何时遍历哪一个list。 好像我这么做是有问题的,应该用Interator<>来做。
Time Complexity - O(n), Space Complexity - O(1)
public class ZigzagIterator { private int length; private int index1 = 0; private int index2 = 0; private List<Integer> list1; private List<Integer> list2; private boolean useList2; public ZigzagIterator(List<Integer> v1, List<Integer> v2) { this.length = v1.size() + v2.size(); list1 = v1; list2 = v2; if(list1.size() == 0) { useList2 = true; } } public int next() { if(!useList2) { if(index2 < list2.size()) { useList2 = true; } return list1.get(index1++); } else { if(index1 < list1.size()) { useList2 = false; } return list2.get(index2++); } } public boolean hasNext() { return (index1 + index2) < length; } } /** * Your ZigzagIterator object will be instantiated and called as such: * ZigzagIterator i = new ZigzagIterator(v1, v2); * while (i.hasNext()) v[f()] = i.next(); */
Reference:
https://leetcode.com/discuss/57961/o-n-time-%26-o-1-space-java-solution
https://leetcode.com/discuss/63037/simple-java-solution-for-k-vector
https://leetcode.com/discuss/58012/short-java-o-1-space
https://leetcode.com/discuss/71857/clean-java-solution-works-for-k-lists
标签:
原文地址:http://www.cnblogs.com/yrbbest/p/5035622.html