[Swift]LeetCode269. 外星人詞典 $ Alien Dictionary

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There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of non-empty words from the dictionary, where words are sorted lexicographically by the rules of this new language. Derive the order of letters in this language.

Example 1:

Input:
[
  "wrt",
  "wrf",
  "er",
  "ett",
  "rftt"
]

Output: 
"wertf"

Example 2:

Input:
[
  "z",
  "x"
]

Output: 
"zx"

Example 3:

Input:
[
  "z",
  "x",
  "z"
] 

Output:  

Explanation: The order is invalid, so return .
""""

Note:

1. You may assume all letters are in lowercase.
2. You may assume that if a is a prefix of b, then a must appear before b in the given dictionary.
3. If the order is invalid, return an empty string.
4. There may be multiple valid order of letters, return any one of them is fine.

---

有一種新的外來語言使用拉丁字母。 可是，您不知道字母之間的順序。 您會從字典中收到一個非空單詞列表，其中的單詞按照新語言的規則按字典順序排序。 導出這種語言的字母順序。

例1：

Input:
[
  "wrt",
  "wrf",
  "er",
  "ett",
  "rftt"
]

Output: "wertf"

例2：

Input:
[
  "z",
  "x"
]

Output: "zx"

例3：

Input:
[
  "z",
  "x",
  "z"
] 

Output:  ""

說明：訂單無效，所以請返回「」。
注意：

一、您能夠假設全部字母都是小寫的。
二、您能夠假設若是a是b的前綴，則a必須出如今給定字典中的b以前。
三、若是訂單無效，則返回空字符串。
四、可能有多個有效的字母順序，返回其中任何一個都沒問題。

---

Solution

 1 class Solution {
 2     func alienOrder(_ words: [String]) -> String {
 3         var st:Set<[Character]> = Set<[Character]>()
 4         var ch:Set<Character> = Set<Character>()
 5         var ins:[Int] = [Int](repeating:0,count:256)
 6         var q:[Character] = [Character]()
 7         var res:String = String()
 8         for a in words
 9         {
10             for c in a
11             {
12                 ch.insert(c)
13             }
14         }
15         for i in 0..<words.count - 1
16         {
17             let mn:Int = min(words[i].count, words[i + 1].count)
18             var j:Int = 0
19             while(j < min(words[i].count, words[i + 1].count))
20             {
21                 if words[i][j] != words[i + 1][j]
22                 {
23                     st.insert([words[i][j], words[i + 1][j]])
24                     break
25                 }
26                 j += 1
27             }
28             if j == mn && words[i].count > words[i + 1].count
29             {
30                 return String()
31             }
32         }
33         for a in st
34         {
35             ins[a[1].ascii] += 1
36         }
37         for a in ch
38         {
39             if ins[a.ascii] == 0
40             {
41                 q.append(a)
42                 res.append(a)
43             }
44         }
45         while(!q.isEmpty)
46         {
47             let c:Character = q.removeFirst()
48             for a in st
49             {
50                 if a[0] == c
51                 {
52                     ins[a[1].ascii] -= 1
53                     if ins[a[1].ascii] == 0
54                     {
55                         q.append(a[1])
56                         res.append(a[1])
57                     }
58                 }
59             }
60         }
61         return res.count == ch.count ? res : String()
62     }
63 }
64 
65 //String擴展
66 extension String {
67     //subscript函數能夠檢索數組中的值
68     //直接按照索引方式截取指定索引的字符
69     subscript (_ i: Int) -> Character {
70         //讀取字符
71         get {return self[index(startIndex, offsetBy: i)]}
72     }
73 }
74 
75 //Character擴展
76 extension Character
77 {
78     //Character轉ASCII整數值(定義小寫爲整數值)
79     var ascii: Int {
80         get {
81             return Int(self.unicodeScalars.first?.value ?? 0)
82         }
83     }
84 }

---

點擊：Playground測試

1 var sol = Solution()
2 print(sol.alienOrder(["wrt", "wrf", "er", "ett","rftt"]))
3 //Print wertf
4 print(sol.alienOrder(["z","x"]))
5 //Print zx
6 print(sol.alienOrder(["z","x","z"]))
7 //Print ""