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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].
给定两个一维向量，实现迭代器交替返回元素。
例如，给定两个一维向量：
v1 = [1, 2] v2 = [3, 4, 5, 6]
通过反复调用next直到hasNext返回false，next返回的元素顺序应该是：[1，3，2，4，5，6]。
后续：如果你得到k 1d向量怎么办？您的代码在这种情况下可以扩展到什么程度？
后续问题澄清-更新（2015-09-18）：
“之字形”顺序没有明确定义，对于k>2的情况不明确。如果“之字形”看起来不正确，请用“循环”替换“之字形”。例如，给定以下输入：
[1,2,3] [4,5,6,7] [8,9]
它应该返回[1,4,8,2,5,9,3,6,7]。
Solution:
1 class ZigzagIterator { 2 var v:[Int] = [Int]() 3 var i:Int = 0 4 init(_ v1:[Int],_ v2:[Int]) 5 { 6 var n1:Int = v1.count 7 var n2:Int = v2.count 8 let n:Int = max(n1, n2) 9 for i in 0..<n 10 { 11 if i < n1 {v.append(v1[i])} 12 if i < n2 {v.append(v2[i])} 13 } 14 } 15 16 func next() -> Int 17 { 18 let num:Int = v[i] 19 i += 1 20 return num 21 } 22 23 func hasNext() -> Bool 24 { 25 return i < v.count 26 } 27 }
点击：Playground测试
1 var zigzag = ZigzagIterator([1, 2],[3, 4, 5, 6]) 2 while(zigzag.hasNext()) 3 { 4 print(zigzag.next()) 5 } 6 //Print 7 /* 8 1 9 3 10 2 11 4 12 5 13 6 14 */