#540 Zigzag Iterator

题目描述：

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

Have you met this question in a real interview? 

Yes

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].

题目思路：

这题的解法有很多。这里用了最土的方法：先将zigzag遍历完的结果存在一个数组中，每次call next()就调用一个element。空间代价是需要一个数组和一个index（始终指向下一个被call的element）。

My code （AC = 44ms）：

class ZigzagIterator {public:    /**     * @param v1 v2 two 1d vectors     */    vector<int> zigzag_list; // the list to store the zigzag elements    int index; // the index indicating which element should be called next time        ZigzagIterator(vector<int>& v1, vector<int>& v2) {        // initialize your data structure here.        int l1 = 0, l2 = 0;        index = 0;                while (l1 < v1.size() && l2 < v2.size()) {            zigzag_list.push_back(v1[l1++]);            zigzag_list.push_back(v2[l2++]);        }                while (l1 < v1.size()) {            zigzag_list.push_back(v1[l1++]);        }                while (l2 < v2.size()) {            zigzag_list.push_back(v2[l2++]);        }    }    int next() {        // Write your code here        return zigzag_list[index++];    }    bool hasNext() {        // Write your code here        return index < zigzag_list.size();    }};/** * Your ZigzagIterator object will be instantiated and called as such: * ZigzagIterator solution(v1, v2); * while (solution.hasNext()) result.push_back(solution.next()); * Ouptut result */