我们用栈同时维护当前字符串和倍数以及要加倍的字符串
当遇到"["时，我们保存当前字符串，即将当前字符 cres 串入栈； 当遇到"]"时，res = cres + 倍数 * 应加倍的字符串
class Solution:
def decodeString(self, s: str) -> str:
stack, res, multi = [], "", 0
for ch in s:
if ch == "[": //保存当前字符串和当前倍数
stack.append([multi, res])
res, multi = "", 0
elif ch == "]":
curm, cres = stack.pop()
res = cres + curm * res //之前保存的字符串 + 应加倍的字符串 * 倍数
elif "0" <= ch <= "9":
multi = multi * 10 + int(ch)
else:
res += ch
return res
标签：multi,ch,--,res,cres,394,字符串,stack From： https://www.cnblogs.com/zk6696/p/17679337.html