s = "3[a]2[bc]", return "aaabcbc". s = "3[a2[c]]", return "accaccacc". s = "2[abc]3[cd]ef", return "abcabccdcdcdef".
1 class Solution(object): 2 def decodeString(self, s): 3 """ 4 :type s: str 5 :rtype: str 6 """ 7 8 stack=[] 9 curNum=0 10 curStr=‘‘ 11 12 for ch in s: 13 if ch==‘[‘: 14 stack.append(curStr) 15 stack.append(curNum) 16 curStr="" 17 curNum=0 18 elif ch==‘]‘: 19 num=stack.pop() 20 preStr=stack.pop() 21 curStr=preStr+num*curStr 22 elif ch.isdigit(): 23 curNum=curNum*10+int(ch) 24 else: 25 curStr+=ch 26 27 return curStr
原文：https://www.cnblogs.com/zijidan/p/12544636.html