Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).

Example:

Input: S = "ADOBECODEBANC", T = "ABC"
Output: "BANC"

Note:

• If there is no such window in S that covers all characters in T, return the empty string ""
  .
• If there is such window, you are guaranteed that there will always be only one unique minimum window in S.

示例：

輸入: S = "ADOBECODEBANC", T = "ABC"
輸出: "BANC"

說明：

• 如果 S 中不存這樣的子串，則返回空字串 ""。
• 如果 S 中存在這樣的子串，我們保證它是唯一的答案。

超出時間限制

 1 class Solution {
 2     func minWindow(_ s: String, _ t: String) -> String {
 3         if t.count > s.count {return String()} 
 4         var res:String = String()
 5         var left = 0, count = 0, minLen = s.count + 1
 6         var tm:[Int] = [Int](repeating:0
 
,count: 128)
 7         var sm:[Int] = [Int](repeating:0,count: 128)
 8         for i in 0..<t.count
 9         {
10             tm[t[i]] += 1
11         }
12         for right in 0..<s.count
13         {
14             let numRight = s[right]
15             if tm[numRight] != 0
16             {
17                  sm[numRight] += 1
18                 if sm[numRight] <= tm[numRight]
19                 {
20                     count += 1
21                 }
22                 while (count == t.count)
23                 {
24                     if right - left + 1 < minLen
25                     {
26                         minLen = right - left + 1
27                         res = s.subString(left, minLen)
28                     }
29                     let numLeft = s[left]
30                     if tm[numLeft] != 0
31                     {
32                         sm[numLeft] -= 1
33                          if sm[numLeft] < tm[numLeft]
34                         {
35                             count -= 1
36                         }
37                     }
38                     left += 1
39                 }
40             }
41         }
42         return res
43     }
44 }
45 
46 extension String {
47     
48     //subscript函式可以檢索陣列中的值
49     //按照索引方式讀取字元
50     subscript (_ num: Int) -> Int {
51         //讀取字元
52         get 
53         {
54             let s = String(self[index(self.startIndex,offsetBy: num)]).unicodeScalars
55             return Int(s[s.startIndex].value) 
56         }
57     }
58     
59     // 擷取字串：指定索引和字元數
60     // - begin: 開始擷取處索引
61     // - count: 擷取的字元數量
62     func subString(_ begin:Int,_ count:Int) -> String {
63         let start = self.index(self.startIndex, offsetBy: max(0, begin))
64         let end = self.index(self.startIndex, offsetBy:  min(self.count, begin + count))
65         return String(self[start..<end]) 
66     }
67 }