[leetcode] 394. Decode String @ ptyhon

最新推荐文章于 2024-06-19 18:22:32 发布

原创最新推荐文章于 2024-06-19 18:22:32 发布 · 95 阅读

0 ·

CC 4.0 BY-SA版权

本文介绍了一种使用堆栈解码重复字符串的高效算法。针对输入的编码字符串，该算法能在O(n)的时间复杂度内返回解码后的字符串。通过实例演示了如何处理如'3[a]2[bc]'等复杂编码。

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