题解 | #大数加法#

    # write code here
    # 判断空的情况
    if s == '':
        return t
    if t == '':
        return s
    
    # 保证s的长度小于等于t
    r = ''
    if len(s) > len(t):
        r = s
        s = t
        t = r

    slength = len(s)
    tlength = len(t)
    
    # 反转字符串方便求和
    s = s[::-1]
    t = t[::-1]
    
    mod = 0 # 进位
    result = ''
    # 首先处理短字符串
    for i in range(slength):
        sum = numb(s[i]) + numb(t[i]) + mod
        mod = 0
        if sum > 9:
            mod = 1
            sum = sum - 10
        result += str(sum)
    # 剩下的长字符串
    for j in range(slength, tlength):
        sum = numb(t[j]) + mod
        mod = 0
        if sum > 9:
            mod = 1
            sum = sum - 10
        result += str(sum)
    # 最后是否还有进位
    if mod == 1:
        result += '1'
    return result[::-1]


def numb(s):
    return ord(s) - 48