0394. Decode String

https://leetcode.com/problems/decode-string

Description

Given an encoded string, return its 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].

Example 1:

**Input:** s = "3[a]2[bc]"
**Output:** "aaabcbc"

Example 2:

**Input:** s = "3[a2[c]]"
**Output:** "accaccacc"

Example 3:

**Input:** s = "2[abc]3[cd]ef"
**Output:** "abcabccdcdcdef"

Example 4:

**Input:** s = "abc3[cd]xyz"
**Output:** "abccdcdcdxyz"

Constraints:

• 1 <= s.length <= 30

• s consists of lowercase English letters, digits, and square brackets '[]'.

• s is guaranteed to be a valid input.

• All the integers in s are in the range [1, 300].

ac1: iterative

class Solution {
    public String decodeString(String s) {
        // edge cases
        if (s == null || s.length() < 3) return s;

        Stack<StringBuilder> stack = new Stack<>();
        stack.push(new StringBuilder());
        Stack<Integer> numstack = new Stack<>();
        int len = s.length();

        for (int i = 0; i < len; i++) {
            char c = s.charAt(i);
            if (c >= '0' && c <= '9') { // meet number
                int n = 0;
                while (i < len && s.charAt(i) >= '0' && s.charAt(i) <= '9') {
                    n = n * 10 + (s.charAt(i) - '0');
                    i++;
                }
                numstack.push(n);
                i--;
            } else if (c == '[') {
                stack.push(new StringBuilder());
            } else if (c == ']') {
                StringBuilder curr = stack.pop();
                String repeatString = curr.toString();
                int k = numstack.pop();
                while (k > 1) {
                    curr.append(repeatString);
                    k--;
                }
                stack.peek().append(curr.toString());
            } else {
                stack.peek().append(c);
            }
        }

        return stack.pop().toString();
    }
}

ac2: recursive

iterative stack == recursive

class Solution {
    int i = 0;
    public String decodeString(String s) {
        int n = 0;
        StringBuilder sb = new StringBuilder();
        while (i < s.length()) {
            if(s.charAt(i) >= '0' && s.charAt(i) <= '9') {
                while (i < s.length() && s.charAt(i) >= '0' && s.charAt(i) <= '9') {
                    n = n * 10 + (s.charAt(i) - '0');
                    i++;
                }
                i++; // skip [
                String next = decodeString(s);
                while ( n > 0) {
                    sb.append(next);
                    n--;
                }
            } else if (s.charAt(i) == ']') {
                i++;
                return sb.toString();
            } else {
                sb.append(s.charAt(i++));
            }
        }

        return sb.toString();
    }
}