[Swift]LeetCode482.密钥格式化|LicenseKeyFormatting

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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:

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

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

 1 class Solution {
 2     func licenseKeyFormatting(_ S: String, _ K: Int) -> String {
 3         var res:[String] = [String]()
 4         for index in S.indices.reversed()
 5         {
 6             if S[index] != "-"
 7             {
 8                 if res.count % (K + 1) == K
 9                 {
10                     res.append("-")
11                 }
12                 res.append(String(S[index]))
13             }
14         }
15         //字符数组转字符串
16         var str:String = String(res.joined(separator: "").reversed())
17         return str.uppercased()
18     }
19 }

132ms

 1 class Solution {
 2     func licenseKeyFormatting(_ S: String, _ K: Int) -> String {
 3         var bufferS: String = ""
 4         var result: String = ""
 5         
 6         if(S.count == 0) { return "" }
 7         
 8         for c in S {
 9             if(String(c) != "-") {
10                 bufferS += (String(c)).uppercased()
11             }
12         }
13         
14         var i: Int = bufferS.count % K == 0 ? K : bufferS.count % K
15         for c in bufferS {
16             if(i == 0) {
17                 result += "-"
18                 i = K
19             }
20             if(i > 0) {
21                 result += String(c)
22                 i -= 1
23             }
24         }
25         
26         return result
27     }
28 }

220ms

 1 class Solution {
 2     func licenseKeyFormatting(_ S: String, _ K: Int) -> String {
 3         let s = S.replacingOccurrences(of: "-", with: "").uppercased()
 4         let chas = [Character](s)
 5         
 6         var res = ""
 7         res.append(String(chas[..<(chas.count%K)]))
 8        
 9         for i in stride(from: chas.count % K, to: chas.count, by: K) {
10             if !res.isEmpty {
11                 res.append("-")
12             }
13             res.append(String(chas[i..<(i+K)]))
14         }
15         
16         return res
17     }
18 }

368ms

 1 class Solution {
 2     func licenseKeyFormatting(_ S: String, _ K: Int) -> String {
 3         var stringArray = S.split(separator: "-").joined(separator: "").map { String($0) }
 4         
 5         var returnString: String = ""
 6         
 7         while(!stringArray.isEmpty) {
 8             let subArray: String = Array(stringArray.suffix(K)).reduce("", +).uppercased()
 9             for _ in 0..<subArray.count {
10                 stringArray.removeLast()
11             }
12             returnString = stringArray.isEmpty ? "\(subArray)" + returnString : "-\(subArray)" + returnString
13             
14         }
15         
16         return returnString
17     }
18 }

1052ms

 1 class Solution {
 2     func licenseKeyFormatting(_ S: String, _ K: Int) -> String {
 3 
 4         var n = 0
 5         for c in S {
 6             if c != "-" {
 7                 n += 1
 8             }
 9         }
10         
11         var num_g = n / K
12         var first = K
13         if n % K > 0 {
14             first = n % K
15             num_g += 1
16         }
17         
18         var res = ""
19         var temp = ""
20         var count_g = 0
21         
22         for c in S {
23             if c != "-" {
24                 temp += String(c).uppercased()
25                 if (count_g == 0 && temp.count == first && count_g != num_g-1) || (count_g < num_g-1 && temp.count == K) {
26                     res += temp + "-"
27                     temp = ""
28                     count_g += 1
29                 } else if (count_g == num_g-1 && temp.count == K) || (count_g == 0 && temp.count == first) {
30                     res += temp
31                     temp = ""
32                     count_g += 1
33                 }
34             }
35         }
36         
37         return res
38     }
39 }