281. Zigzag Iterator

• Difficulty: Medium

• Topics: Design

• Similar Questions:

  • Binary Search Tree Iterator

  • Flatten 2D Vector

  • Peeking Iterator

  • Flatten Nested List Iterator

Problem:

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

Example:

Input:
v1 = [1,2]
v2 = [3,4,5,6] 

Output: [1,3,2,4,5,6]

Explanation: 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:
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:

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

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

Solutions:

class ZigzagIterator {
public:
    ZigzagIterator(vector<int>& v1, vector<int>& v2) {
        this->v1 = v1;
        this->v2 = v2;
    }

    int next() {
        if (pos1 == v1.size()) {
            return v2[pos2++];
        }

        if (pos2 == v2.size()) {
            return v1[pos1++];
        }

        if ((pos1 + pos2) % 2 == 0) {
            return v1[pos1++];
        } else {
            return v2[pos2++];
        }
    }

    bool hasNext() {
        return pos1 < v1.size() || pos2 < v2.size();
    }
private:
    int pos1 = 0;
    int pos2 = 0;
    vector<int> v1;
    vector<int> v2;
};

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