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

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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:

class ZigzagIterator {
    var v:[Int] = [Int]()
    var i:Int = 0
    init(_ v1:[Int],_ v2:[Int])
    {
        var n1:Int = v1.count
        var n2:Int = v2.count
        let n:Int = max(n1, n2)
        for i in 0..<n
        {
            if i < n1 {v.append(v1[i])}
            if i < n2 {v.append(v2[i])}
        }
    }
    
    func next() -> Int
    {
        let num:Int = v[i]
        i += 1
        return num
    }
    
    func hasNext() -> Bool
    {
        return i < v.count
    }
}

---

点击：Playground测试

var zigzag = ZigzagIterator([1, 2],[3, 4, 5, 6])
while(zigzag.hasNext())
{
    print(zigzag.next())
}
//Print 
/*
1
3
2
4
5
6
*/