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.

 

思路：判断一个数字字符应该转为$a-j$的字符还是应该和后一位数字字符结合转为$j-z$，应该看第三位字符，如果是$#$，则需要将其和后一位字符结合。

C++:

 1 class Solution {
 2 public:
 3     string freqAlphabets(string s) {
 4         string ans;
 5         for (int i = 0; i < s.length(); ++i) {
 6             int num = s[i] - '0';
 7             if (i + 2 < s.length() && (s[i + 2] == '#')) {
 8                 num = num * 10 + (s[i + 1] - '0');
 9                 i += 2;
10             }
11             ans.push_back('a' + num - 1);
12         }
13         return ans;
14     }
15 };

python3:

1 import re
2 class Solution:
3     def freqAlphabets(self, s: str) -> str:
4         return ''.join([chr(int(i[:2]) + 96) for i in re.findall(r'\d\d#|\d', s)])