• 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].

     1 public class ZigzagIterator {
     2     private int index1 = 0, index2 = 0;
     3     private List<Integer> l1;
     4     private List<Integer> l2;
     5     
     6     public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
     7         l1 = v1;
     8         l2 = v2;
     9     }
    10 
    11     public int next() {
    12         if (index1 < l1.size()) {
    13             if (index2 < l2.size()) {
    14                 if (index2 < index1) return l2.get(index2++);
    15             }
    16             return l1.get(index1++);
    17         } else return l2.get(index2++);
    18     }
    19 
    20     public boolean hasNext() {
    21         return index1 < l1.size() || index2 < l2.size();
    22     }
    23 }
    24 
    25 /**
    26  * Your ZigzagIterator object will be instantiated and called as such:
    27  * ZigzagIterator i = new ZigzagIterator(v1, v2);
    28  * while (i.hasNext()) v[f()] = i.next();
    29  */
  • 相关阅读:
    Elasticsearch与kibana的单机安装
    PS 设计带斑点图案的背景
    中国传统纹样简略
    js 在非module中引用module里的变量和函数
    ssh 使用pem秘钥文件登录
    jquery sortable 使用注意事项
    ES6中的函数、对象定义
    file-loader返回object Module 路径的问题
    Louvain 论文笔记
    基于Docker方式的LNMP环境搭建
  • 原文地址:https://www.cnblogs.com/amazingzoe/p/6416414.html
Copyright © 2020-2023  润新知