[Swift]LeetCode482. 密钥格式化 | License Key Formatting

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You are given a license key represented as a string S which consists only alphanumeric character and dashes. The string is separated into N+1 groups by N dashes.

Given a number K, we would want to reformat the strings such that each group contains exactly K characters, except for the first group which could be shorter than K, but still must contain at least one character. Furthermore, there must be a dash inserted between two groups and all lowercase letters should be converted to uppercase.

Given a non-empty string S and a number K, format the string according to the rules described above.

Example 1:

Input: S = "5F3Z-2e-9-w", K = 4

Output: "5F3Z-2E9W"

Explanation: The string S has been split into two parts, each part has 4 characters.
Note that the two extra dashes are not needed and can be removed.

Example 2:

Input: S = "2-5g-3-J", K = 2

Output: "2-5G-3J"

Explanation: The string S has been split into three parts, each part has 2 characters except the first part as it could be shorter as mentioned above.

Note:

1. The length of string S will not exceed 12,000, and K is a positive integer.
2. String S consists only of alphanumerical characters (a-z and/or A-Z and/or 0-9) and dashes(-).
3. String S is non-empty.

---

给定一个密钥字符串S，只包含字母，数字以及 '-'（破折号）。N 个 '-' 将字符串分成了 N+1 组。给定一个数字 K，重新格式化字符串，除了第一个分组以外，每个分组要包含 K 个字符，第一个分组至少要包含 1 个字符。两个分组之间用 '-'（破折号）隔开，并且将所有的小写字母转换为大写字母。

给定非空字符串 S 和数字 K，按照上面描述的规则进行格式化。

示例 1：

输入：S = "5F3Z-2e-9-w", K = 4

输出："5F3Z-2E9W"

解释：字符串 S 被分成了两个部分，每部分 4 个字符；
     注意，两个额外的破折号需要删掉。

示例 2：

输入：S = "2-5g-3-J", K = 2

输出："2-5G-3J"

解释：字符串 S 被分成了 3 个部分，按照前面的规则描述，第一部分的字符可以少于给定的数量，其余部分皆为 2 个字符。

提示:

1. S 的长度不超过 12,000，K 为正整数
2. S 只包含字母数字（a-z，A-Z，0-9）以及破折号'-'
3. S 非空

---

class Solution {
    func licenseKeyFormatting(_ S: String, _ K: Int) -> String {
        var res:[String] = [String]()
        for index in S.indices.reversed()
        {
            if S[index] != "-"
            {
                if res.count % (K + 1) == K
                {
                    res.append("-")
                }
                res.append(String(S[index]))
            }
        }
        //字符数组转字符串
        var str:String = String(res.joined(separator: "").reversed())
        return str.uppercased()
    }
}

---

132ms

class Solution {
    func licenseKeyFormatting(_ S: String, _ K: Int) -> String {
        var bufferS: String = ""
        var result: String = ""
        
        if(S.count == 0) { return "" }
        
        for c in S {
            if(String(c) != "-") {
                bufferS += (String(c)).uppercased()
            }
        }
        
        var i: Int = bufferS.count % K == 0 ? K : bufferS.count % K
        for c in bufferS {
            if(i == 0) {
                result += "-"
                i = K
            }
            if(i > 0) {
                result += String(c)
                i -= 1
            }
        }
        
        return result
    }
}

---

220ms

class Solution {
    func licenseKeyFormatting(_ S: String, _ K: Int) -> String {
        let s = S.replacingOccurrences(of: "-", with: "").uppercased()
        let chas = [Character](s)
        
        var res = ""
        res.append(String(chas[..<(chas.count%K)]))
       
        for i in stride(from: chas.count % K, to: chas.count, by: K) {
            if !res.isEmpty {
                res.append("-")
            }
            res.append(String(chas[i..<(i+K)]))
        }
        
        return res
    }
}

---

368ms

class Solution {
    func licenseKeyFormatting(_ S: String, _ K: Int) -> String {
        var stringArray = S.split(separator: "-").joined(separator: "").map { String($0) }
        
        var returnString: String = ""
        
        while(!stringArray.isEmpty) {
            let subArray: String = Array(stringArray.suffix(K)).reduce("", +).uppercased()
            for _ in 0..<subArray.count {
                stringArray.removeLast()
            }
            returnString = stringArray.isEmpty ? "(subArray)" + returnString : "-(subArray)" + returnString
            
        }
        
        return returnString
    }
}

---

1052ms

class Solution {
    func licenseKeyFormatting(_ S: String, _ K: Int) -> String {

        var n = 0
        for c in S {
            if c != "-" {
                n += 1
            }
        }
        
        var num_g = n / K
        var first = K
        if n % K > 0 {
            first = n % K
            num_g += 1
        }
        
        var res = ""
        var temp = ""
        var count_g = 0
        
        for c in S {
            if c != "-" {
                temp += String(c).uppercased()
                if (count_g == 0 && temp.count == first && count_g != num_g-1) || (count_g < num_g-1 && temp.count == K) {
                    res += temp + "-"
                    temp = ""
                    count_g += 1
                } else if (count_g == num_g-1 && temp.count == K) || (count_g == 0 && temp.count == first) {
                    res += temp
                    temp = ""
                    count_g += 1
                }
            }
        }
        
        return res
    }
}