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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。
1 class Solution { 2 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 }