[Leetcode] 281. Zigzag Iterator 解题报告

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] .

思路：

我自己开始的实现用了5个变量：1个bool变量表示下一个数是否是在v1中取，还有4个迭代器分别表示v1和v2的当前位置和结束位置。虽然测试可以通过，但是代码非常冗余。后来看到网上用队列的思路，十分巧妙，这里顺便实现以下。

代码：

class ZigzagIterator {
public:
    ZigzagIterator(vector<int>& v1, vector<int>& v2) {
        if (v1.size() > 0) {
            q.push(make_pair(v1.begin(), v1.end()));
        }
        if (v2.size() > 0) {
            q.push(make_pair(v2.begin(), v2.end()));
        }
    }

    int next() {
        int ret = *q.front().first;
        auto it1 = q.front().first, it2 = q.front().second;
        if (it1 + 1 != it2) {
            q.push(make_pair(it1 + 1, it2));
        }
        q.pop();
        return ret;
    }

    bool hasNext() {
        return !q.empty();
    }
private:
    queue<pair<vector<int>::iterator, vector<int>::iterator>> q;
};

/**
 * Your ZigzagIterator object will be instantiated and called as such:
 * ZigzagIterator i(v1, v2);
 * while (i.hasNext()) cout << i.next();
 */