[Swift]LeetCode186. 翻轉字符串中的單詞 II $ Reverse Words in a String II

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公衆號：山青詠芝（shanqingyongzhi）
➤博客園地址：山青詠芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：http://www.javashuo.com/article/p-pruclynf-md.html 
➤若是連接不是山青詠芝的博客園地址，則多是爬取做者的文章。
➤原文已修改更新！強烈建議點擊原文地址閱讀！支持做者！支持原創！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

Given an input string, reverse the string word by word. A word is defined as a sequence of non-space characters.
The input string does not contain leading or trailing spaces and the words are always separated by a single space.
For example,
Given s = "the sky is blue",
return "blue is sky the".
Could you do it in-place without allocating extra space?

---

給定輸入字符串，逐字反轉字符串。單詞被定義爲非空格字符序列。

輸入字符串不包含前導或後綴空格，而且單詞老是由一個空格分隔。

例如，

給定  s = "the sky is blue",

返回 "blue is sky the".。

您能在不分配額外空間的狀況下就地完成嗎？

---

 1 class Solution {
 2     func reverseWords(_ s: inout String){
 3         var left:Int = 0
 4         for i in 0...s.count
 5         {
 6             if i == s.count || s[i] == " "
 7             {
 8                 reverse(&s, left, i - 1)
 9                 left = i + 1
10             }
11         }
12         reverse(&s, 0, s.count - 1)
13     }
14     
15     func reverse(_ s: inout String,_ left:Int,_ right:Int)
16     {
17         var left = left
18         var right = right
19         while (left < right)
20         {
21             var t:Character = s[left]
22             s[left] = s[right]
23             s[right] = t
24             left += 1
25             right -= 1
26         }
27     }
28 }
29 
30 extension String {        
31     //subscript函數能夠檢索數組中的值
32     //直接按照索引方式截取指定索引的字符
33     subscript (_ i: Int) -> Character {
34         //讀取字符
35         get {return self[index(startIndex, offsetBy: i)]}
36         
37         //修改字符
38         set
39         {
40             var str:String = self
41             var index = str.index(startIndex, offsetBy: i)
42             str.remove(at: index)
43             str.insert(newValue, at: index)
44             self = str
45         }
46     }
47 }