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".

java

class Solution {    public String decodeString(String s) {        if (s == null || s.length() == 0) {            return "";        }        Stack<Object> stack = new Stack<>();        int num = 0;        for (char c : s.toCharArray()) {            if (Character.isDigit(c)) {                num = num * 10 + c - '0';            } else if (c == '[') {                stack.push(Integer.valueOf(num));                num = 0;            } else if (c == ']') {                String str = util(stack);                int count = (int)stack.pop();                for (int i = 0; i < count; i++) {                    stack.push(str);                }            } else {                stack.push(String.valueOf(c));            }        }        return util(stack);    }    private String util(Stack<Object> stack) {        StringBuffer sb = new StringBuffer();        Stack<String> stack1 = new Stack<>();        while (!stack.empty() && stack.peek() instanceof String) {            stack1.push((String)stack.pop());        }        while (!stack1.empty()) {            sb.append(stack1.pop());        }        return sb.toString();    }}

python

class Solution:    """    @param: s: an expression includes numbers, letters and brackets    @return: a string    """    def expressionExpand(self, s):        # write your code here        if s is None or len(s) == 0:            return ""        num = 0        stack = []        for element in s:            if element.isdigit():                num = num * 10 + int(element);            elif element is '[':                stack.append(num)                num = 0            elif element is ']':                s = self.util(stack)                count = stack.pop()                for i in range(count):                    stack.append(s)            else:                stack.append(element)        return self.util(stack)            def util(self, stack):        s = []        while len(stack) != 0 and isinstance(stack[-1], str):            s.append(stack.pop())        s.reverse()        return "".join(s)