Decode String

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".

Now it comes with layer by layer, we know stack should be the best choice. For this problem, we keep two stacks, one for the numbers, the other for the letters. When we see a number, we push it to the number stack. When we see a "[", we add all current layer letters to the stack. When we see a "]", we append current layer string with correct repeating times to last layer string.

public String decodeString(String s) {
        if (s == null || s.length() == 0) {
            return s;
        }
        Stack repeatingTimes = new Stack<>();
        Stack layer = new Stack<>();
        String rst = "";
        int pos = 0;
        int len = s.length();
        while (pos < len) {
            char c = s.charAt(pos);
            if (Character.isDigit(c)) {
                int curr = pos;
                while (Character.isDigit(s.charAt(curr))) {
                    curr++;
                }
                repeatingTimes.push(Integer.parseInt(s.substring(pos, curr)));
                pos = curr;
            } else if (Character.isLetter(c)) {
                rst += c;
                pos++;
            } else if (c == '[') {
                layer.push(rst);
                rst = "";
                pos++;
            } else if (c == ']') {
                StringBuilder builder = new StringBuilder();
                int times = repeatingTimes.pop();
                while (times > 0) {
                    builder.append(rst);
                    times--;
                }
                rst = layer.pop() + builder.toString();
                pos++;
            } else {
                return "";
            }
        }
        return rst;
    }