刷题之路径总和Ⅲ（leetcode）

a = '123'

def concatenate_within_limit(b, new_string):
    # 计算新字符串与a的长度之和
    a = b
    total_length = len(a) + len(new_string)

    # 如果长度超过1024，从前面删除足够的字符
    if total_length > 5:
        diff = total_length - 5
        a = a[diff:] + new_string  # 删除前diff个字符，并拼接新字符串
    else:
        a += new_string  # 长度未超过1024，直接拼接新字符串

    # 返回处理后的a
    return a

for i in range(4,10):
    print(concatenate_within_limit(a, str(i)))