Decode String

Problem

Given an encoded string, return it's decoded string.

The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer.

You may assume that the input string is always valid; No extra white spaces, square brackets are well-formed, etc.

Furthermore, you may assume that the original data does not contain any digits and that digits are only for those repeat numbers, k. For example, there won't be input like 3a or 2[4].

Examples:

s = "3[a]2[bc]", return "aaabcbc".
s = "3[a2[c]]", return "accaccacc".
s = "2[abc]3[cd]ef", return "abcabccdcdcdef".

Solution

Typical Parsing Nested Elements Solution Using Stacks

public class Solution {
    public String decodeString(String s) {
        Stack<Integer> intS = new Stack<>();
        Stack<StringBuilder> sbS = new Stack<>();
        StringBuilder sb = new StringBuilder();
        int k = 0;
        for (char ch : s.toCharArray()) {
            if (Character.isDigit(ch)) k = k * 10 + (ch - '0');
            else if (ch == '[') {
                intS.push(k);
                k = 0;
                sbS.push(sb);
                sb = new StringBuilder();
            }
            else if (ch == ']') {
                StringBuilder temp = sb;
                sb = sbS.pop();
                k = intS.pop();
                while (k-- > 0) sb.append(temp.toString());
                k = 0; //k is -1 here, hence we need to set it to 0
            }
            else sb.append(ch);
        }
        return sb.toString();
    }
}

Analysis

Since this is a nested parsing, we should use stack
If reach [, we need to store the repeated times k and current sb and set sb to new one
If reach ], we need to pop both stack to get k and previous sb, then add previous sb to current sb
At the end, return sb.toString()