281. 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].
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
最新文章
- JavaScript实现DOM对象选择器
- Android中使用ShareSDK集成分享功能
- mysql常用函数
- 通过Canvas + JS 实现简易时钟实战
- C语言中printf的规范输出
- iOS开发学习概述及知识整理
- CSS预处理器Sass、LESS 和 Stylus
- Oracle12C相关
- mm/mmap.c
- R&S学习笔记（一）
- 一、IOS运行原理
- 每天一个linux命令（35）：killall命令
- 每天一个linux命令（21）：tar命令
- vi 每日练习
- jaspersoft 5.6.0 相关问题
- 安卓常用 widget
- [HDOJ2572]终曲
- PyQt5之窗口类型
- python之vscode配置开发调试环境
- ECMAScript 6 -- 数组的解构赋值