【leetcode】1309. Decrypt String from Alphabet to Integer Mapping

题目如下：

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.

解题思路：遍历s，判断s[i]后面的第二个元素是否是'#'。如果不是，s[i]转换成对应字母；如果是，s[i]和s[i+1]一起转换成对应字母。

代码如下：

class Solution(object):
    def freqAlphabets(self, s):
        """
        :type s: str
        :rtype: str
        """
        res = ''
        inx = 0
        while inx < len(s):
            if s[inx] == '#':
                inx += 1
            elif inx + 2 < len(s) and s[inx+2] == '#':
                res += chr(ord('a') + int(s[inx] + s[inx+1]) - 1)
                inx += 2
            else:
                res += chr(ord('a') + int(s[inx]) - 1)
                inx += 1
        return res