哈希表实现通讯录功能

以下是使用哈希表实现通讯录功能的Python代码：

class Contact:
    def __init__(self, name, phone):
        self.name = name
        self.phone = phone

class HashTable:
    def __init__(self, size):
        self.size = size
        self.table = [[] for _ in range(self.size)]

    def _hash_function(self, name):
        return sum([ord(c) for c in name]) % self.size

    def insert(self, contact):
        hash_key = self._hash_function(contact.name)
        bucket = self.table[hash_key]
        for i, c in enumerate(bucket):
            if c.name == contact.name:
                bucket[i] = contact
                return
        bucket.append(contact)

    def delete(self, name):
        hash_key = self._hash_function(name)
        bucket = self.table[hash_key]
        for i, c in enumerate(bucket):
            if c.name == name:
                del bucket[i]
                return

    def search(self, name):
        hash_key = self._hash_function(name)
        bucket = self.table[hash_key]
        for c in bucket:
            if c.name == name:
                return c.phone
        return None

def menu():
    print("1. 添加联系人")
    print("2. 删除联系人")
    print("3. 查找联系人")
    print("4. 退出")
    return int(input("请选择操作："))

if __name__ == '__main__':
    contacts = HashTable(10)
    while True:
        choice = menu()
        if choice == 1:
            name = input("请输入姓名：")
            phone = input("请输入电话号码：")
            contact = Contact(name, phone)
            contacts.insert(contact)
        elif choice == 2:
            name = input("请输入要删除的联系人姓名：")
            contacts.delete(name)
        elif choice == 3:
            name = input("请输入要查找的联系人姓名：")
            phone = contacts.search(name)
            if phone:
                print(f"{name}的电话号码是{phone}")
            else:
                print(f"{name}不存在")
        elif choice == 4:
            break
        else:
            print("无效的选择")