标签：may encoding for lse des only ant else ali
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].
Examples:
s = "3[a]2[bc]", return "aaabcbc". s = "3[a2[c]]", return "accaccacc". s = "2[abc]3[cd]ef", return "abcabccdcdcdef".
Solution 1: Use 2 stacks
class Solution: def decodeString(self, s: str) -> str: char_stack = [] num_stack = [] res, num, index = ‘‘, 0, 0 while index < len(s): char = s[index] if char.isdigit(): num = 0 while index < len(s) and s[index].isdigit(): num = 10 * num + int(s[index]) index += 1 num_stack.append(num) elif char == ‘[‘: char_stack.append(res) res = ‘‘ index += 1 elif char == ‘]‘: cur_char_lst = list(char_stack.pop()) times = num_stack.pop() for _ in range(times): cur_char_lst.append(res) res = ‘‘.join(cur_char_lst) index += 1 else: res += char index += 1 return res
solution 2: Use 1 stack
class Solution: def decodeString(self, s: str) -> str: stack = [] cur_string, cur_num = ‘‘, 0 for c in s: if c.isdigit(): cur_num = 10 * cur_num + int(c) elif c == ‘[‘: stack.append(cur_string) stack.append(cur_num) cur_string = ‘‘ cur_num = 0 elif c == ‘]‘: prev_num = stack.pop() prev_string = stack.pop() cur_string = prev_string + prev_num * cur_string else: cur_string += c return cur_string
标签：may encoding for lse des only ant else ali
原文地址：https://www.cnblogs.com/xuanlu/p/11797621.html
