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?微信公眾號(hào)：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(gè)一維向量，實(shí)現(xiàn)迭代器交替返回元素。

例如，給定兩個(gè)一維向量：

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

通過反復(fù)調(diào)用next直到hasNext返回false，next返回的元素順序應(yīng)該是：[1，3，2，4，5，6]。

后續(xù)：如果你得到k 1d向量怎么辦？您的代碼在這種情況下可以擴(kuò)展到什么程度？

后續(xù)問題澄清-更新（2015-09-18）：

“之字形”順序沒有明確定義，對(duì)于k>2的情況不明確。如果“之字形”看起來不正確，請(qǐng)用“循環(huán)”替換“之字形”。例如，給定以下輸入：

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

它應(yīng)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(diǎn)擊：Playground測(cè)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 */

?

轉(zhuǎn)載于:https://www.cnblogs.com/strengthen/p/10686001.html

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