[leetcode281]Zigzag Iterator

Question:

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].

分析：

实现一个迭代器，此迭代器交叉访问两个数组。

很简单，先访问v1，判断v1是否还有数据，若有数据则输出，否则直接输出v2剩余所有元素。

再访问v2，同样判断v2是否还有数据，若有数据则输出，否则直接输出v1剩余所有数据。

依次交叉来。

代码如下：

class ZigzagIterator {public:    ZigzagIterator(vector<int>& v1, vector<int>& v2) {        bs[0] = v1.begin(), bs[1] = v2.begin();        es[0] = v1.end(), es[1] = v2.end();        p = 0;    }    int next() {        int elem;        if (bs[0] == es[0])              elem = *bs[1]++;        else                if (bs[1] == es[1])                     elem = *bs[0]++;               else {                        elem = *bs[p]++;                        p = (p + 1) % 2;                }        return elem;    }    bool hasNext() {        return bs[0] != es[0] || bs[1] != es[1];    }private:    int p;    vector<int>::iterator bs[2], es[2];};