【leetcode】394. Decode String【M】【45ms】

最新推荐文章于 2021-08-01 01:02:51 发布

原创最新推荐文章于 2021-08-01 01:02:51 发布 · 682 阅读

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#leetcode #leetcode题解 #python #string #栈

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本文介绍了一种使用栈数据结构解码特定格式字符串的算法。该算法能解析形如 k[encoded_string] 的字符串，其中 encoded_string 被重复 k 次。通过逐字符遍历并利用栈来维护当前结构，实现了解码过程。

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 thatk 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 or2[4].

Examples:

s = "3[a]2[bc]", return "aaabcbc".
s = "3[a2[c]]", return "accaccacc".
s = "2[abc]3[cd]ef", return "abcabccdcdcdef".

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类似于计算表达式，什么逆波兰，前缀表达式之类的

用一个栈，来维护当前的结构

这个还是很经典的

而且这样一想，整个题目就简单了

class Solution(object):
    def decodeString(self, s):
        i = 0
        stack = []
        temp = ''

        while i < len(s):
            num = 0
            #print s[i:],'+++',stack
            while i + 1 < len(s) and s[i].isdigit():
                num = num * 10 + int(s[i])
                i += 1
            #print num
            if num != 0:
                stack += num,

            if s[i] == '[':
                stack += '['
                i += 1
                continue

            if s[i] == ']':
                i += 1
                temp = ''
                top = stack.pop()

                while top != '[':
                    #print top
                    temp += top
                    #print '~temp:',temp
                    top = stack.pop()
                num = stack.pop()
                #print num
                temp *= int(num)
                #print temp
                stack += [temp]
                continue
                # break

            if i < len(s):
                stack += s[i],
            i += 1


        res = ''
        #print stack
        for i in stack:
            res += i[::-1]
        return res