281. Zigzag Iterator

最新推荐文章于 2025-09-11 18:44:42 发布

转载最新推荐文章于 2025-09-11 18:44:42 发布 · 72 阅读

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#数据结构与算法

本文介绍了一种名为ZigzagIterator的交错迭代器实现方法，该方法可以将两个一维向量的元素交替返回。此外，还探讨了如何将此方法扩展到处理多个一维向量的情况，并提供了一个遵循循环顺序的例子。

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();
 */