题解 | #大数加法#

#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 计算两个数之和
# @param s string字符串 表示第一个整数
# @param t string字符串 表示第二个整数
# @return string字符串
#
class Solution:
    def solve(self, s: str, t: str) -> str:
        def addss(s1, s2, ls):
            m, n = len(s), len(t)
            i = j = flag = 0
            while i < m and j < n:
                temp = int(s[i]) + int(t[j]) + flag
                if 0 <= temp <= 9:
                    ls.append(temp)
                    flag = 0
                else:
                    temp = str(temp)
                    ls.append(temp[1])
                    flag = 1
                i += 1
                j += 1
                if i == m and j == n and flag == 1:
                    ls.append(1)
            return i, j, flag, ls

        def adds(p, flag, ss, res):
            while p < len(ss):
                if flag == 0:
                    res.append(ss[p:])
                    break
                else:
                    temp = int(ss[p]) + flag
                    if 0 <= temp <= 9:
                        res.append(ss[p])
                        flag = 0
                    else:
                        temp = str(temp)
                        res.append(temp[1])
                        flag = 1
                p += 1
                if p == len(ss) and flag == 1:
                    res.append(1)
            return res

        s, t = s[::-1], t[::-1]
        res = []
        # 另一个字符串为空
        if s == "":
            return t
        if t == "":
            return s
        # 相加
        i, j, flag, res = addss(s, t, res)
        # 处理剩余部分
        res = adds(i, flag, s, res)
        res = adds(j, flag, t, res)
        # 对顺序进行处理
        res.reverse()
        return ''.join(str(x) for x in res)