Formatted question description: https://leetcode.ca/all/1309.html

1309. Decrypt String from Alphabet to Integer Mapping (Easy)

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.

Related Topics:
String

Solution 1.

// OJ: https://leetcode.com/problems/decrypt-string-from-alphabet-to-integer-mapping/

// Time: O(N)
// Space: O(1)
class Solution {
public:
    string freqAlphabets(string s) {
        string ans;
        for (int i = 0, N = s.size(); i < N; ++i) {
            if (i + 2 < N && s[i + 2] == '#') {
                ans += 'a' + (s[i] - '0') * 10 + (s[i + 1] - '0') - 1;
                i += 2;
            } else ans += 'a' + s[i] - '0' - 1;
        }
        return ans;
    }
};

Java

class Solution {
    public String freqAlphabets(String s) {
        Stack<Character> stack = new Stack<Character>();
        int length = s.length();
        int index = length - 1;
        while (index >= 0) {
            char c = s.charAt(index);
            if (c == '#') {
                int num = Integer.parseInt(s.substring(index - 2, index));
                char letter = (char) (num - 1 + 'a');
                stack.push(letter);
                index -= 3;
            } else {
                char letter = (char) (c - '1' + 'a');
                stack.push(letter);
                index--;
            }
        }
        StringBuffer sb = new StringBuffer();
        while (!stack.isEmpty())
            sb.append(stack.pop());
        return sb.toString();
    }
}

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