实现IP地址聚合计算问题

Question

实现IP地址聚合计算。
支持多种输入格式。
至少实现2种IP地址聚合算法。
参考如下：
http://apps.neu.edu.cn/netaggr/

Answer 1

用python实现：

import ipaddress

# 定义IP地址聚合函数
def ip_aggregate(ip_list, algorithm='equal_length'):
    # 将IP地址字符串转换为IPv4Address对象
    ip_objects = [ipaddress.IPv4Network(ip) for ip in ip_list]
    # 根据算法排序
    if algorithm == 'equal_length':
        ip_objects = sorted(ip_objects, key=lambda x: (x.prefixlen, int(x.network_address)))
    elif algorithm == 'longest_prefix':
        ip_objects = sorted(ip_objects, key=lambda x: (int(x.network_address), x.prefixlen))
    else:
        raise ValueError('Unknown algorithm: %s' % algorithm)
    # 初始化聚合结果列表
    result = []
    # 依次比较IP地址段进行聚合
    current = ip_objects[0]
    for ip in ip_objects[1:]:
        if algorithm == 'equal_length':
            if current.prefixlen == ip.prefixlen and \
               int(ip.network_address) == (int(current.broadcast_address) + 1):
                # 可以聚合
                current = ipaddress.IPv4Network(current.network_address, current.prefixlen - 1)
            else:
                # 不能聚合，输出当前CIDR块并更新current
                result.append(str(current))
                current = ip
        elif algorithm == 'longest_prefix':
            prefixlen = current.prefixlen
            while not ip.subnet_of(current) and prefixlen > 0:
                prefixlen -= 1
                current = ipaddress.IPv4Network(current.network_address, prefixlen)
            # 输出当前CIDR块并更新current
            result.append(str(current))
            current = current.supernet()
    # 输出最后一个CIDR块
    result.append(str(current))
    return result

# 测试IP地址聚合函数
ip_list = ['192.168.0.0/24', '192.168.2.0/24', '192.168.1.0/24']
result1 = ip_aggregate(ip_list, algorithm='equal_length')
result2 = ip_aggregate(ip_list, algorithm='longest_prefix')
print('Equal length algorithm result:', result1)
print('Longest prefix algorithm result:', result2)