lc281-Zigzag Iterator

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Given two 1d vectors, implement an iterator to return their elements alternately.

For example, given two 1d vectors:

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2
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:

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2
3
[1,2,3]
[4,5,6,7]
[8,9]

It should return [1,4,8,2,5,9,3,6,7].

用cur代表当前位置，cur % k代表要读哪个vector，cur/k是那个vector要读取的index。

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class ZigzagIterator {
    int m, n;
    int cur = -1;
    vector<int> v1, v2;
public:
    ZigzagIterator(vector<int>& v1, vector<int>& v2) {
        m = v1.size();
        n = v2.size();
        this -> v1 = v1;
        this -> v2 = v2;
    }
    int next() {
        cur++;
        if (cur / 2 < min(m, n)) {
            if (cur % 2) return v2[cur/2];
            else return v1[cur/2];
        }
        else if (m < n) {
            return v2[cur-m];
        }
        else return v1[cur-n];
    }
    bool hasNext() {
        return cur+1 < m+n;
    }
};
/**
 * Your ZigzagIterator object will be instantiated and called as such:
 * ZigzagIterator i(v1, v2);
 * while (i.hasNext()) cout << i.next();
 */