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.

class Solution {
    public String freqAlphabets(String s) {
        StringBuilder sb = new StringBuilder();
        if(s.length() == 0 || s == null || s.charAt(0) == '#') return "";
        for(int i = 0; i < s.length(); i++){
            if((i+2) < s.length() && s.charAt(i+2) == '#'){
                sb.append((char) ('a' + Integer.parseInt(s.substring(i, i+2)) - 1));
                i += 2;
            }
            else{
                sb.append((char) ('a' + Integer.parseInt(s.substring(i, i+1)) - 1));
            }
        }
        return sb.toString();
    }
}

可以直接做，先判断第三位是不是#，如果是就append前两位数对应的字母，记得要i = i + 2，总共应该是个+3跳过整个字母的对应，因为for循环中还有个+1所以这里+2就好了

如果第三位不是#，就说明是10位一下，方法照做就好，最后从头+1跳过就可以。

StringBuilder比String要好