[Swift]LeetCode1100. 长度为 K 的无重复字符子串 | Find K-Length Substrings With No Repeated Characters

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Given a string S, return the number of substrings of length K with no repeated characters.

Example 1:

Input: S = "havefunonleetcode", K = 5 Output: 6 Explanation: There are 6 substrings they are : 'havef','avefu','vefun','efuno','etcod','tcode'. 

Example 2:

Input: S = "home", K = 5 Output: 0 Explanation: Notice K can be larger than the length of S. In this case is not possible to find any substring.

Note:

1. 1 <= S.length <= 10^4
2. All characters of S are lowercase English letters.
3. 1 <= K <= 10^4

---

给你一个字符串 S，找出全部长度为 K 且不含重复字符的子串，请你返回所有知足要求的子串的 数目。 

示例 1：

输入：S = "havefunonleetcode", K = 5
输出：6
解释：
这里有 6 个知足题意的子串，分别是：'havef','avefu','vefun','efuno','etcod','tcode'。

示例 2：

输入：S = "home", K = 5
输出：0
解释：
注意：K 可能会大于 S 的长度。在这种状况下，就没法找到任何长度为 K 的子串。 

提示：

1. 1 <= S.length <= 10^4
2. S 中的全部字符均为小写英文字母
3. 1 <= K <= 10^4

---

44ms

 1 class Solution {
 2     func numKLenSubstrNoRepeats(_ S: String, _ K: Int) -> Int {
 3         var ans:Int = 0
 4         let n:Int = S.count
 5         let S:[Int] = Array(S).map{$0.ascii}
 6         for i in 0..<n
 7         {
 8             var freq:[Int] = [Int](repeating:0,count:26)
 9             var j:Int = i
10             var len:Int = 0
11             while(j < n)
12             {
13                 //a:97
14                 if freq[S[j] - 97] != 0 {break}
15                 freq[S[j] - 97] += 1
16                 len += 1
17                 j += 1
18                 if len == K
19                 {
20                     ans += 1
21                 }
22             }
23         }
24         return ans
25     }
26 }
27 
28 //Character扩展
29 extension Character
30 {
31     //Character转ASCII整数值(定义小写为整数值)
32     var ascii: Int {
33         get {
34             return Int(self.unicodeScalars.first?.value ?? 0)
35         }
36     }
37 }