[leetcode] 281. Zigzag Iterator 解题报告

题目链接:https://leetcode.com/problems/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] .

思路: 将两个一维数组中的数都先保存起来, 然后设置一个索引位来返回next, 如果已经到了尽头则返回false.

代码如下:

class ZigzagIterator {
public:
    ZigzagIterator(vector<int>& v1, vector<int>& v2) {
        index = 0;
        hash.reserve(v1.size()+v2.size());
        int k = 0;
        while(k < v1.size() || k < v2.size())
        {
            if(k < v1.size()) hash.push_back(v1[k]);
            if(k < v2.size()) hash.push_back(v2[k]);
            k++;
        }
    }

    int next() {
        return hash[index++];
    }

    bool hasNext() {
        return index < hash.size();
    }
private:
    int index;
    vector<int> hash;
};

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