394. 字符串解码

题目链接	394. 字符串解码
思路	字符串模拟；出现相同子问题，可以使用递归或者栈解决
题解链接	字符串解码（辅助栈法 / 递归法，清晰图解）
关键点	栈：需要存储（重复次数，当前字符串）；递归：需要范围 内嵌字符串及结束位置
时间复杂度	\(O(n)\)
空间复杂度	\(O(n)\)

代码实现（栈）：

class Solution:
    def decodeString(self, s: str) -> str:
        stk = []
        num = 0
        answer = ""
        for ch in s:
            if ch.isdigit():
                num = num * 10 + int(ch)
            elif ch == "[":
                stk.append((answer, num))
                answer, num = "", 0
            elif ch == "]":
                prev_str, cur_times = stk.pop()
                answer = prev_str + cur_times * answer
            else:
                answer += ch
        return answer

代码实现（递归）：

class Solution:
    def decodeString(self, s: str) -> str:
        def decode(i):
            answer, n = "", 0
            while i < len(s):
                if s[i].isdigit():
                    n = n * 10 + int(s[i])
                elif s[i] == '[':
                    i, nested = decode(i+1)
                    answer += n * nested
                    n = 0
                elif s[i] == ']':
                    return i, answer
                else:
                    answer += s[i]
                i += 1
            return answer
        return decode(0)