394. Decode String

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; there are 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 will not be input like 3a or 2[4].

The test cases are generated so that the length of the output will never exceed 105.

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"

Solution:

class Solution {
    public String decodeString(String s) {

        /*
        0 1 2 3 4 5 6 7 8
        3 [ a ] 2 [ b c ]
                        i

        countStack: [ 2
        resultStack: [ "aaa"
        currentString: "bc"
        k = 0 = 0 * 10 + (3 -'0') = 3
        k = 0 * 10 + (2 -'0') =2
        k = 0

        2

        aaabcbc


        */
        Stack<Integer> countStack = new Stack<>();

        Stack<StringBuilder> resultStack = new Stack<>();
        StringBuilder currentString = new StringBuilder();
        int k = 0;

        for (char ch : s.toCharArray()){
            if (Character.isDigit(ch)){
                // If it's a digit, build the number (k can be more than one digit long)
                k = k * 10 + (ch - '0');
            }else if (ch == '['){
                // Push the current k and currentString into their respective stacks
                countStack.push(k);
                resultStack.push(currentString);
                // Reset k and currentString for the new section inside the brackets
                k = 0;
                currentString = new StringBuilder();
            }else if (ch == ']'){
                // Pop the last count and last string from the stacks
                int repeatTimes = countStack.pop();
                StringBuilder decodedString = resultStack.pop();
                // Append the currentString repeated 'repeatTimes' to the decodedString
                for (int i = 0; i < repeatTimes; i++){
                    decodedString.append(currentString);
                }

                // Update teh currentString to the decoded string
                currentString = decodedString;
            }else{
                // If it's a character, just add it to the currentString
                currentString.append(ch);
            }
        }

        return currentString.toString();
    }
}
// TC: O(kn)
// SC:

This problem is hard for me.

Solution process: use an example to follow step then you will understand a litte

And then write 3 times++

Record the video to express.