Given two 1d vectors, implement an iterator to return their elements alternately.
Example:
Input: v1 = [1,2] v2 = [3,4,5,6] Output:[1,3,2,4,5,6] Explanation:
By calling next repeatedly until hasNext returnsfalse
, 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:
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:
Input:
[1,2,3]
[4,5,6,7]
[8,9]
Output: [1,4,8,2,5,9,3,6,7]
.
A:
这里,最先错的地方,就是 iterator, 不需要拷贝原有内容的。
这个比较傻逼。是因为自己以前没有看过真正的代码实现。
class ZigzagIterator { public: ZigzagIterator(vector<int>& v1, vector<int>& v2) { s1 = v1.begin(); s2 = v2.begin(); e1 = v1.end(); e2 = v2.end(); isFirst = true; } int next() { int res=0; if(s1==e1){ res = *s2; s2++; }else if(s2==e2){ res = *s1; s1++; }else{ if(isFirst){ res = *s1; isFirst = ! isFirst; s1++; }else{ res = *s2; isFirst = ! isFirst; s2++; } } return res; } bool hasNext() { return s1 !=e1 || s2 != e2; } private: vector<int>::iterator s1,s2,e1,e2; bool isFirst; }; /** * Your ZigzagIterator object will be instantiated and called as such: * ZigzagIterator i(v1, v2); * while (i.hasNext()) cout << i.next(); */
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