Given a string s, return all the palindromic permutations (without duplicates) of it. Return an empty list if no palindromic permutation could be form.

For example:

Given s = "aabb", return ["abba", "baab"].

Given s = "abc", return [].

Hint:

1. If a palindromic permutation exists, we just need to generate the first half of the string.
2. To generate all distinct permutations of a (half of) string, use a similar approach from: Permutations II or Next Permutation.

給定的字符串S，回報所有的palindromic置換（沒有duplicates）的信息。返回的列表，如果沒有空palindromic -可以排列形式。

例如：

給定s = "aabb",返回的 ["abba", "baab"].

給定 s = "abc",，返回 [ ]。

提示：

1. 如果一palindromic排列是否，我們只需要兩個Generate的第一個字符串。
2. Generate置換成不同的兩個全（半）的字符串，用相似的方法：Permutations II 或者 Next Permutation.。

 1 class Solution {
 2     func generatePalindromes(_ s:String) -> [String]{
 3         var res:Set<String> = Set<String>()
 4         var m:[Character:Int] = [Character:Int]()
 5         var t:String = ""
 

 6         var mid:String = ""
 7         for a in s.characters
 8         {
 9             m[a] = 1
10         }
11         for (key, val) in m
12         {
13             if val % 2 == 1
14             {
15                 mid.append(key)
16             }
17             var str:String = String()
18             for i in 0..<val/2
19             {
20                 str.append(key)
21             }
22             t += str
23             if mid.count > 1
24             {
25                 return []
26             }
27         }
28         permute(&t, 0, mid,&res);
29         return Array(res)
30     }
31     
32     func permute(_ t:inout String,_ start:Int,_ mid:String,_ res:inout Set<String>)
33     {
34         if start >= t.count
35         {
36             let str:String = String(t.reversed())
37             res.insert(t + mid + str)
38         }
39         for i in start..<t.count
40         {
41             if i != start && t[i] == t[start]
42             {
43                 continue
44             }
45             var temp:Character
46             temp = t[start]
47             t[start] = t[i]
48             t[i] = temp
49             
50             permute(&t, start + 1, mid, &res)
51             
52             temp = t[start]
53             t[start] = t[i]
54             t[i] = temp
55         }
56     }
57 }
58 
59 extension String {        
60     //subscript函數可以檢索數組中的值
61     //直接按照索引方式截取指定索引的字符
62     subscript (_ i: Int) -> Character {
63         //讀取字符
64         get {return self[index(startIndex, offsetBy: i)]}
65         
66         //修改字符
67         set
68         {
69             var str:String = self
70             var index = str.index(startIndex, offsetBy: i)
71             str.remove(at: index)
72             str.insert(newValue, at: index)
73             self = str
74         }
75     }
76 }

[Swift]LeetCode267.回文全排列 II $ Palindrome Permutation II