[Swift]LeetCode467. 环绕字符串中唯一的子字符串 | Unique Substrings in Wraparound String

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Consider the string s to be the infinite wraparound string of "abcdefghijklmnopqrstuvwxyz", so s will look like this: "...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....".

Now we have another string p. Your job is to find out how many unique non-empty substrings of p are present in s. In particular, your input is the string p and you need to output the number of different non-empty substrings of p in the string s.

Note: p consists of only lowercase English letters and the size of p might be over 10000.

Example 1:

Input: "a"
Output: 1

Explanation: Only the substring "a" of string "a" is in the string s. 

Example 2:

Input: "cac"
Output: 2
Explanation: There are two substrings "a", "c" of string "cac" in the string s. 

Example 3:

Input: "zab"
Output: 6
Explanation: There are six substrings "z", "a", "b", "za", "ab", "zab" of string "zab" in the string s.

---

把字符串 s 看作是“abcdefghijklmnopqrstuvwxyz”的无限环绕字符串，所以 s 看起来是这样的："...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....". 

现在我们有了另一个字符串 p 。你需要的是找出 s 中有多少个唯一的 p的非空子串，尤其是当你的输入是字符串 p ，你需要输出字符串 s 中 p的不同的非空子串的数目。 

注意: p 仅由小写的英文字母组成，p 的大小可能超过 10000。 

示例 1:

输入: "a"
输出: 1
解释: 字符串 S 中只有一个"a"子字符。 

示例 2:

输入: "cac"
输出: 2
解释: 字符串 S 中的字符串“cac”只有两个子串“a”、“c”。 

示例 3:

输入: "zab"
输出: 6
解释: 在字符串 S 中有六个子串“z”、“a”、“b”、“za”、“ab”、“zab”。

---

8236ms

class Solution {
    func findSubstringInWraproundString(_ p: String) -> Int {
        var cnt:[Int] = [Int](repeating:0,count:26)
        var res:Int = 0
        var len:Int = 0
        for i in 0..<p.count
        {
            var cur:Int = p[i].ascii - 97
            if i > 0 && p[i - 1].ascii != (cur + 26 - 1) % 26 + 97
            {
                len = 0
            }
            len += 1
            if len > cnt[cur]
            {
                res += len - cnt[cur]
                cnt[cur] = len
            }
            
        }
        return res
    }
}

extension String {        
    //subscript函数可以检索数组中的值
    //直接按照索引方式截取指定索引的字符
    subscript (_ i: Int) -> Character {
        //读取字符
        get {return self[index(startIndex, offsetBy: i)]}
    }
}

extension Character  
{  
  //属性：ASCII整数值(定义小写为整数值)
   var ascii: Int {
        get {
            let s = String(self).unicodeScalars
            return Int(s[s.startIndex].value)
        }
    }
}

---

Memory Usage: 4194kb

class Solution {
    func findSubstringInWraproundString(_ p: String) -> Int {
        var count: [Character: Int] = [ : ]
        let s = Array("abcdefghijklmnopqrstuvwxyz")
        let p = Array(p)
        for c in s {
            count[c] = 0
        }
        
        var maxLen = 0
        
        for i in 0 ..< p.count {
            if i > 0 && (p[i].unicodeScalars.first!.value == p[i - 1].unicodeScalars.first!.value + 1 || p[i - 1].unicodeScalars.first!.value == p[i].unicodeScalars.first!.value + 25) {
                maxLen += 1
            } else {
                maxLen = 1
            }
            count[p[i]] = max(count[p[i]]!, maxLen)
        }
        
        return count.values.reduce(0, { $0 + $1 })
    }
}