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].
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首先這題其實不是之前的zigzag,已經clarification了。說白了這題就是按列輸出所有的number。跟之前那道iterator有相似之處。
題目給了兩個vector, 但follow up question其實希望algorithm可以適用於k個vector。所以類似BFS, 記得cc150說: first think about using queue when talked about BFS.
class ZigzagIterator {
public:
ZigzagIterator(vector<int>& v1, vector<int>& v2) {
if(!v1.empty()) q.push(make_pair(v1.begin(), v1.end()));
if(!v2.empty()) q.push(make_pair(v2.begin(), v2.end()));
}
int next() {
vector<int>:: iterator curB = q.front().first;
vector<int>:: iterator curE = q.front().second;
q.pop();
if(curB + 1 != curE) q.push(make_pair(curB+1, curE));
return (*curB);
}
bool hasNext() {
return !q.empty();
}
private:
queue<pair<vector<int>:: iterator, vector<int>:: iterator> > q;
};
/**
* Your ZigzagIterator object will be instantiated and called as such:
* ZigzagIterator i(v1, v2);
* while (i.hasNext()) cout << i.next();
*/