[LeetCode281]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].
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首先这题其实不是之前的zigzag，已经clarification了。说白了这题就是按列输出所有的number。跟之前那道iterator有相似之处。

题目给了两个vector， 但follow up question其实希望algorithm可以适用于k个vector。所以类似BFS, 记得cc150说： first think about using queue when talked about BFS.

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

    int next() {
        vector<int>:: iterator curB = q.front().first;
        vector<int>:: iterator curE = q.front().second;
        q.pop();
        if(curB + 1 != curE) q.push(make_pair(curB+1, curE));
        return (*curB);
    }

    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(); */