大数加法 -- 每日一题01

题目：以字符串的形式读入两个数字，编写一个函数计算它们的和，以字符串形式返回。示例:输入:"1","99"输出:"100"

数据范围：，字符串仅由'0'~‘9’构成

要求：时间复杂度 

思路：想起来小学学数学，如99+11，先取末尾的9和1，9+1=10，进1取0，再取下一个9和1加刚刚进位上来的1，即

9+1+1=11，取1进1，之后就没什么取的了，之前三步取上来的 1 1 0，就是结果110

代码：

class Solution:
    def solve(self, s: str, t: str) -> str:
        # write code here
        s_list = list(s)
        t_list = list(t)
        next = 0
        res = []
        while len(s_list) > 0 and len(t_list) > 0:
            pop_num = int(s_list.pop()) + int(t_list.pop()) + next
            if pop_num >= 10:
                next = 1
            else:
                next = 0
            pop_num = str(pop_num)
            res.append(pop_num[len(pop_num)-1])

        if len(s_list) == 0 and len(t_list) == 0:
            if next == 1:
                res.append('1')
            res.reverse()
            return ''.join(res)

        if len(t_list) == 0:
            while len(s_list) > 0:
                add_num = next + int(s_list.pop())
                if add_num >= 10:
                    next = 1
                else:
                    next = 0
                add_num = str(add_num)
                res.append(add_num[len(add_num)-1])

        if len(s_list) == 0:
            while len(t_list) > 0:
                add_num = next + int(t_list.pop())
                if add_num >= 10:
                    next = 1
                else:
                    next = 0
                add_num = str(add_num)
                res.append(add_num[len(add_num)-1])
        if next == 1:
            res.append('1')
        res.reverse()
        return ''.join(res)