# 394. Decode String 字符串解码

@TOC

## # 题目描述

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".

## # 解题方法

### # 栈

curstring保存的是出栈操作完成后的字符串。

class Solution(object):
def decodeString(self, s):
"""
:type s: str
:rtype: str
"""
curnum = 0
curstring = ''
stack = []
for char in s:
if char == '[':
stack.append(curstring)
stack.append(curnum)
curstring = ''
curnum = 0
elif char == ']':
prenum = stack.pop()
prestring = stack.pop()
curstring = prestring + prenum * curstring
elif char.isdigit():
curnum = curnum * 10 + int(char)
else:
curstring += char
return curstring

## # 日期

2018 年 2 月 17 日 2019 年 1 月 2 日 —— 2019年开刷