LeetCode 1309. 解码字母到整数映射

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"

class Solution:
    def freqAlphabets(self, s: str) -> str:
        #寻找所有#的开头索引
        import re
        twoNumIndex = [m.start()-2 for m in re.finditer("#", s)]
        for i in twoNumIndex[::-1]:
            char = chr(96+int(s[i:i+2]))
            s = s[:i] + char + s[i+3:]
        for j in range(len(s)):
            if s[j].isdigit():
                char = chr(96 + int(s[j]))
                s = s[:j] + char + s[j+1:]
        return s