题目
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; 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].
解题思路
使用栈，每次遇到左括号之后，将当前的数和字母压栈，遇到右括号出栈，并将当前的字符重复栈中数字的次数，并与之前的字符串进行连接
代码
class Solution:
def decodeString(self, s: str) -> str:
stack = []
cur_num = cur_str = ''
for c in s:
if c == '[':
stack.append(cur_str)
stack.append(int(cur_num))
cur_num = cur_str = ''
elif c == ']':
num = stack.pop()
prev_str = stack.pop()
cur_str = prev_str + cur_str * num
elif c.isdigit():
cur_num += c
else:
cur_str += c
return cur_str
