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 = "xyz3[a12[cb]]", return "....."

Solution1：two Stacks (count, result)

思路：到數字時，將原來的res存入res_stack，數字存入count_stack，res重置 ，繼續處理后序。碰到]時，將res_stack.pop() + res * count_stack.pop() times，繼續處理后序。
Time Complexity: O(N) Space Complexity: O(N)

Solution2：Recursive

參考：https://discuss.leetcode.com/topic/57228/0ms-simple-c-solution
Time Complexity: O(N) Space Complexity: O(N) 遞歸緩存

Solution1 Code：

public class Solution {
    public String decodeString(String s) {
        String res = "";
        Stack<Integer> countStack = new Stack<>();
        Stack<String> resStack = new Stack<>();
        int idx = 0;
        while (idx < s.length()) {
            if (Character.isDigit(s.charAt(idx))) {
                int count = 0;
                while (Character.isDigit(s.charAt(idx))) {
                    count = 10 * count + (s.charAt(idx) - '0');
                    idx++;
                }
                countStack.push(count);
            }
            else if (s.charAt(idx) == '[') {
                resStack.push(res);
                res = "";
                idx++;
            }
            else if (s.charAt(idx) == ']') {
                StringBuilder temp = new StringBuilder (resStack.pop());
                int repeatTimes = countStack.pop();
                for (int i = 0; i < repeatTimes; i++) {
                    temp.append(res);
                }
                res = temp.toString();
                idx++;
            }
            else {
                res += s.charAt(idx++);
            }
        }
        return res;
    }
}

Solution2 Code：

public class Solution {
    public String decodeString(String s) {
        StringBuilder result = decode(s, new int[] {0});
        return result.toString();
    }
    
    // start: //so that it can be changed. (like reference)
    private StringBuilder decode(String s, int[] start) {
        StringBuilder result = new StringBuilder();
        
        while(start[0] < s.length() && s.charAt(start[0]) != ']') {
            if(Character.isDigit(s.charAt(start[0]))) {
                int count = 0;
                while (Character.isDigit(s.charAt(start[0]))) {
                    count = 10 * count + (s.charAt(start[0]) - '0');
                    start[0]++;
                }
                
                start[0]++; // ']'
                StringBuilder post = decode(s, start);
                start[0]++; // ']'
                
                while(count > 0) {
                    result.append(post);
                    count--;
                }
                
            }
            else {
                result.append(s.charAt(start[0]));
                start[0]++;
            }
        }
        return result;
    }
}

Round1

class Solution {
    public String decodeString(String s) {
        
        Deque<String> stack_str = new ArrayDeque<>();
        Deque<Integer> stack_count = new ArrayDeque<>();
        
        int i = 0;
        
        StringBuilder str = new StringBuilder("");
        while(i < s.length()) {  
            if(Character.isDigit(s.charAt(i))) {
                int num = 0;
                while(Character.isDigit(s.charAt(i))) {
                    num = num * 10 + s.charAt(i) - '0';
                    i++;
                }
                stack_count.push(num);
            }
            
            else if(s.charAt(i) == '[') {
                stack_str.push(str.toString());
                str = new StringBuilder("");
                i++;
            }
            else if(s.charAt(i) == ']') {
                StringBuilder prev = new StringBuilder(stack_str.pop());
                int times = stack_count.pop();
                for(int t = 0; t < times; t++) {
                    prev.append(str);
                }
                str = prev;
                i++;
            }
            else {
                // regular string
                str.append(s.charAt(i));
                i++;
            }
        }
        return str.toString();
    }
}