Medium
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]
.
Example 1:
Input: s = “3[a]2[bc]”
Output: “aaabcbc”
Example 2:
Input: s = “3[a2[c]]”
Output: “accaccacc”
Example 3:
Input: s = “2[abc]3[cd]ef”
Output: “abcabccdcdcdef”
Example 4:
Input: s = “abc3[cd]xyz”
Output: “abccdcdcdxyz”
Constraints:
1 <= s.length <= 30
s
consists of lowercase English letters, digits, and square brackets '[]'
.s
is guaranteed to be a valid input.s
are in the range [1, 300]
.class Solution:
def __init__(self):
self.i = 0
def decodeString(self, s: str) -> str:
count = 0
sb = []
while self.i < len(s):
c = s[self.i]
self.i += 1
if c.isalpha():
sb.append(c)
elif c.isdigit():
count = count * 10 + int(c)
elif c == ']':
break
elif c == '[':
# sub problem
repeat = self.decodeString(s)
sb.append(repeat * count)
count = 0
return ''.join(sb)