[Swift]LeetCode281. 之字形迭代器 $ Zigzag Iterator

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➤微信公衆號：山青詠芝（shanqingyongzhi）
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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].

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給定兩個一維向量，實現迭代器交替返回元素。

例如，給定兩個一維向量：

v1 = [1, 2]
v2 = [3, 4, 5, 6]

經過反覆調用next直到hasNext返回false，next返回的元素順序應該是：[1，3，2，4，5，6]。

後續：若是你獲得k 1d向量怎麼辦？您的代碼在這種狀況下能夠擴展到什麼程度？

後續問題澄清-更新（2015-09-18）：

「之字形」順序沒有明肯定義，對於k>2的狀況不明確。若是「之字形」看起來不正確，請用「循環」替換「之字形」。例如，給定如下輸入：

[1,2,3]
[4,5,6,7]
[8,9]

它應該返回[1,4,8,2,5,9,3,6,7]。

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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 }

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點擊：Playground測試

 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 */