[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，找出所有长度为 K 且不含重复字符的子串，请你返回全部满足要求的子串的 数目。 

示例 1：

输入：S = "havefunonleetcode", K = 5
输出：6
解释：
这里有 6 个满足题意的子串，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。在这种情况下，就无法找到任何长度为 K 的子串。 

提示：

1. 1 <= S.length <= 10^4
2. S 中的所有字符均为小写英文字母
3. 1 <= K <= 10^4

---

44ms

class Solution {
    func numKLenSubstrNoRepeats(_ S: String, _ K: Int) -> Int {
        var ans:Int = 0
        let n:Int = S.count
        let S:[Int] = Array(S).map{$0.ascii}
        for i in 0..<n
        {
            var freq:[Int] = [Int](repeating:0,count:26)
            var j:Int = i
            var len:Int = 0
            while(j < n)
            {
                //a:97
                if freq[S[j] - 97] != 0 {break}
                freq[S[j] - 97] += 1
                len += 1
                j += 1
                if len == K
                {
                    ans += 1
                }
            }
        }
        return ans
    }
}

//Character扩展
extension Character
{
    //Character转ASCII整数值(定义小写为整数值)
    var ascii: Int {
        get {
            return Int(self.unicodeScalars.first?.value ?? 0)
        }
    }
}