[Swift]LeetCode1309. 解码字母到整数映射 | Decrypt String from Alphabet to Integer Mapping

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Given a string s formed by digits ('0' - '9') and '#' . We want to map s to English lowercase characters as follows:

Characters ('a' to 'i') are represented by ('1' to '9') respectively.
Characters ('j' to 'z') are represented by ('10#' to '26#') respectively.

Return the string formed after mapping.

It's guaranteed that a unique mapping will always exist.

Example 1:

Input: s = "10#11#12"
Output: "jkab"
Explanation: "j" -> "10#" , "k" -> "11#" , "a" -> "1" , "b" -> "2".

Example 2:

Input: s = "1326#"
Output: "acz"

Example 3:

Input: s = "25#"
Output: "y"

Example 4:

Input: s = "12345678910#11#12#13#14#15#16#17#18#19#20#21#22#23#24#25#26#"
Output: "abcdefghijklmnopqrstuvwxyz"

Constraints:

1 <= s.length <= 1000
s[i] only contains digits letters ('0'-'9') and '#' letter.
s will be valid string such that mapping is always possible.

给你一个字符串 s，它由数字（'0' - '9'）和 '#' 组成。我们希望按下述规则将 s 映射为一些小写英文字符：

字符（'a' - 'i'）分别用（'1' - '9'）表示。
字符（'j' - 'z'）分别用（'10#' - '26#'）表示。

返回映射之后形成的新字符串。

题目数据保证映射始终唯一。

示例 1：

输入：s = "10#11#12"
输出："jkab"
解释："j" -> "10#" , "k" -> "11#" , "a" -> "1" , "b" -> "2".

示例 2：

输入：s = "1326#"
输出："acz"

示例 3：

输入：s = "25#"
输出："y"

示例 4：

输入：s = "12345678910#11#12#13#14#15#16#17#18#19#20#21#22#23#24#25#26#"
输出："abcdefghijklmnopqrstuvwxyz"

提示：

1 <= s.length <= 1000
s[i] 只包含数字（'0'-'9'）和 '#' 字符。
s 是映射始终存在的有效字符串。

Runtime: 4 ms

Memory Usage: 21.1 MB

 1 class Solution {
 2     func freqAlphabets(_ s: String) -> String {
 3         let s:[Character] = Array(s)
 4         var res:String = String()
 5         var i:Int = 0
 6         while(i < s.count)
 7         {
 8             if i < s.count - 2 && s[i + 2] == "#"
 9             {
10                 //"j":106 , "0": 48 , "1":49
11                 let num:Int =  106 +  (Int(s[i].asciiValue!)  - 49) * 10 + Int(s[i + 1].asciiValue!) - 48
12                 res.append(num.ASCII)
13                 i += 2
14             }
15             else
16             {
17                 //a:97 , "1":49
18                 let num:Int =  97 + Int(s[i].asciiValue! ) -  49
19                 res.append(num.ASCII)
20             }
21             i += 1
22         }
23         return res
24     }
25 }
26 
27 //Int扩展
28 extension Int
29 {
30     //Int转Character,ASCII值(定义大写为字符值)
31     var ASCII:Character
32     {
33         get {return Character(UnicodeScalar(self)!)}
34     }
35 }

Runtime: 16 ms

Memory Usage: 21.3 MB

 1  class Solution {
 2     private let chars = Array<Character>(" abcdefghijklmnopqrstuvwxyz")
 3     func freqAlphabets(_ s: String) -> String {
 4         var ans = ""
 5         var sCopy = s
 6         while !sCopy.isEmpty {
 7             if let index = sCopy.firstIndex(of: "#"), sCopy.distance(from: sCopy.startIndex, to: index) == 2 {
 8                 let indexStr = String(sCopy[sCopy.startIndex...sCopy.index(after: sCopy.startIndex) ])
 9                 sCopy.removeFirst(3)
10                 ans.append(chars[Int(indexStr)!])
11             } else {
12                 ans.append(chars[Int("\(sCopy.removeFirst())")!])
13             }
14         }
15         return ans
16     }
17  }