组合两个表(C++)

+-------------+---------+
| 列名         | 类型     |
+-------------+---------+
| PersonId    | int     |
| FirstName   | varchar |
| LastName    | varchar |
+-------------+---------+
personId 是该表的主键列。
该表包含一些人的 ID 和他们的姓和名的信息。

+-------------+---------+
| 列名         | 类型    |
+-------------+---------+
| AddressId   | int     |
| PersonId    | int     |
| City        | varchar |
| State       | varchar |
+-------------+---------+
addressId 是该表的主键列。
该表的每一行都包含一个 ID = PersonId 的人的城市和州的信息。

输入: 
Person表:
+----------+----------+-----------+
| personId | lastName | firstName |
+----------+----------+-----------+
| 1        | Wang     | Allen     |
| 2        | Alice    | Bob       |
+----------+----------+-----------+
Address表:
+-----------+----------+---------------+------------+
| addressId | personId | city          | state      |
+-----------+----------+---------------+------------+
| 1         | 2        | New York City | New York   |
| 2         | 3        | Leetcode      | California |
+-----------+----------+---------------+------------+
输出: 
+-----------+----------+---------------+----------+
| firstName | lastName | city          | state    |
+-----------+----------+---------------+----------+
| Allen     | Wang     | Null          | Null     |
| Bob       | Alice    | New York City | New York |
+-----------+----------+---------------+----------+
解释: 
地址表中没有 personId = 1 的地址，所以它们的城市和州返回 null。
addressId = 1 包含了 personId = 2 的地址信息。

# Write your MySQL query statement below
select FirstName, LastName, City, state
from Person left join Address
on Person.PersonId = Address.PersonId;