LeetCode -- 394. 字符串解码(栈处理字符串问题)

 

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