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  • [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  }
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  • 原文地址:https://www.cnblogs.com/strengthen/p/12151953.html
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