1309. Decrypt String from Alphabet to Integer Mapping

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.

遍历一次，按照规则进行转换即可。

class DecryptStringFromAlphabetToIntegerMapping : public Solution {
public:
    void Exec() {
        auto res = freqAlphabets("10#11#12");
    }
    string freqAlphabets(string s) {
        int start_at = 0;
        string res;
        for (int i = 1; i < s.size(); i++) {
            if (s[i] == '#') {
                for (int j = start_at; j <= i - 3; j++) {
                    res.push_back('a' + (s[j] - '1'));
                }
                if (i - 1 - start_at + 1 >= 2) {
                    int num = (s[i - 2] - '0') * 10 + (s[i - 1] - '0');
                    res.push_back('j' + num - 10);
                }
                start_at = i + 1;
            } else if (i == s.size() - 1) {
                for (int j = start_at; j <= i; j++) {
                    res.push_back('a' + (s[j] - '1'));
                }
            }
        }
        return res;
    }
};