LeetCode 281. Zigzag Iterator

#include <vector>#include <iostream>using namespace std;/*  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]  return [1, 3, 2, 4, 5, 6]*/class ZigzagIterator {private:  int index;  vector<int> zigzagOrder;  public ZigzagIterator(vector<int>& v1, vector<int>& v2) {    int i = 0, j = 0;    bool turn = true;    while(i < v1.size() && j < v2.size()) {      if(turn) {        zigzagOrder.push_back(v1[i++]);      } else {        zigzagOrder.push_back(v2[j++]);      }      turn = !turn;    }    while(i < v1.size()) {      zigzagOrder.push_back(v1[i++]);    }    while(j < v2.size()) {      zigzagOrder.push_back(v2[j++]);    }  }  bool hasNext() {    return index < zigzagOrder.size();  }  int next() {    return zigzagOrder[index++];  }

Follow Up: What if you are given k 1d vectors? How well can your code be extended to such cases?

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].

This follow up can be easily solved using deque structure.

  vector<int> zigzagOrder;  int index;  // k vectors  struct iteraterStartAndEnd {    vector<int>::iterator begin;    vector<int>::iterator end;  }  public zigzagIterator(vector< vector<int> >& nums) {    deque<iteraterStartAndEnd> iterators;    for(int i = 0; i < nums.size(); ++i) {      iteraterStartAndEnd tmp;      tmp.begin = nums[i].begin();      tmp.end = nums[i].end();      iterators.push_back(tmp);    }    while(!iterators.empty()) {      auto tmp = iterators.front();      if(tmp.begin != tmp.end()) {        zigzagOrder.push_back(*(tmp.begin));      }      iterators.pop_front();      tmp.begin++;      if(tmp.begin != tmp.end) {        iterators.push_back(tmp);      }    }  }