[leetcode] 394. Decode String @ ptyhon

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”.

解法

使用堆栈, 当遇到正括号时将已读取的字符串和数字进栈, 遇到反括号时出栈.
Time: O(n)
Space: O(1)

class Solution:
    def decodeString(self, s):
        """
        :type s: str
        :rtype: str
        """
        stack = []
        cur_num = 0
        cur_str = ''
        for c in s:
            if c.isdigit():
                cur_num = 10*cur_num + int(c)
            elif c == '[':
                stack.append(cur_str)
                stack.append(cur_num)
                cur_str = ''
                cur_num = 0
            elif c == ']':
                num = stack.pop()
                pre_str = stack.pop()
                cur_str = pre_str + cur_str*num
            else:
                cur_str += c
                
        return cur_str