Given a pattern and a string str, find if strfollows the same pattern.

Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty substring in str.

Example 1:

Input: pattern = "abab", str = "redblueredblue"
Output: true

Example 2:

Input: pattern = "aaaa"
 
, str = "asdasdasdasd"
Output: false

Example 3:

Input: pattern = "aabb", str = "xyzabcxzyabc"
Output: false

Notes:
　　You may assume both pattern and str contains only lowercase letters.

給定一個模式和一個字符串str，找出str是否遵循相同的模式。

這裏follow意味著完全匹配，這樣在模式中的字母和str中的非空子字符串之間就有一個雙射。

例1：

　　輸入：pattern=“abab”，str=“redblueredblue”

　　輸出：true

例2：

　　輸入：pattern=“aaaa”，str=“asdasdasd”

　　輸出：false

例3：

　　輸入：pattern=“aabb”，str=“xyzabcxzyabc”

　　輸出：false

註意：

　　您可以假定pattern和str都只包含小寫字母。

Solution:

 1 class Solution {
 2     var m:[Character:String] = [Character:String]()
 3
 
     var s:Set<String> = Set<String>()
 4     func wordPatternMatch(_ pattern:String,_ str:String) -> Bool {
 5         if pattern.isEmpty {return str.isEmpty}
 6         var arrChar:[Character] = Array( pattern)
 7         if m[arrChar[0]] != nil
 8         {
 9             var t:String = m[arrChar[0]]!
10             if t.count > str.count || str.subString(0, t.count) != t
11             {
12                 return false
13             }
14             if wordPatternMatch(pattern.subString(1), str.subString(t.count))
15             {
16                 return true
17             }
18         }
19         else
20         {
21             for i in 1...str.count
22             {
23                 if s.contains(str.subString(0, i)) {continue}
24                 m[arrChar[0]] = str.subString(0, i)
25                 s.insert(str.subString(0, i))
26                 if wordPatternMatch(pattern.subString(1), str.subString(i))
27                 {
28                     return true
29                 }
30                 m[arrChar[0]] = nil
31                 s.remove(str.subString(0, i))
32             }
33         }
34         return false        
35     }
36 }
37 
38 extension String {
39     // 截取字符串：從index到結束處
40     // - Parameter index: 開始索引
41     // - Returns: 子字符串
42     func subString(_ index: Int) -> String {
43         let theIndex = self.index(self.endIndex, offsetBy: index - self.count)
44         return String(self[theIndex..<endIndex])
45     }
46     
47     // 截取字符串：指定索引和字符數
48     // - begin: 開始截取處索引
49     // - count: 截取的字符數量
50     func subString(_ begin:Int,_ count:Int) -> String {
51         let start = self.index(self.startIndex, offsetBy: max(0, begin))
52         let end = self.index(self.startIndex, offsetBy:  min(self.count, begin + count))
53         return String(self[start..<end]) 
54     }
55 }

點擊：Playground測試

1 var sol = Solution()
2 print(sol.wordPatternMatch("abab","redblueredblue"))
3 //Print true
4 print(sol.wordPatternMatch("aaaa","asdasdasdasd"))
5 //Print false
6 print(sol.wordPatternMatch("aabb","xyzabcxzyabc"))
7 //Print false

[Swift]LeetCode291. 詞語模式之二 $ Word Pattern II