JavaScript如何将嵌套的json对象数据设置为单个嵌套映射

我想将嵌套的JSON数据设置成嵌套的映射并迭代它。考虑下面的示例JSON，我想将FixNew、LaSTNED、地址对象、地址1对象值设置成单个嵌套的映射。还可以如何迭代它来从地址对象城市字段值获得值。 请提供更好的解决方案。 示例json：

 [{
    "firstName": "Jihad",
    "lastName": "Saladin",
    "address": {
      "street": "12 Beaver Court",
      "city": "Snowmass",
      "state": "CO",
      "zip": "81615"
    },
    "address1": {
      "street": "16 Vail Rd",
      "city": "Vail",
      "state": "CO",
      "zip": "81657"
    }
  }]

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如果要合并所有地址，可以执行以下操作

const people = [{
  "firstName": "Jihad",
  "lastName": "Saladin",
  "address": {
    "street": "12 Beaver Court",
    "city": "Snowmass",
    "state": "CO",
    "zip": "81615"
  },
  "address1": {
    "street": "16 Vail Rd",
    "city": "Vail",
    "state": "CO",
    "zip": "81657"
  }
}]
const newPeople = people.map(person => {
  const addresses = [person.address]
  for (let i = 1; person['address' + i]; i++) {
    addresses.push(person['address' + i])
  }
  return {
    firstName: person.firstName,
    lastName: person.lastName,
    addresses
  }
})
console.log(JSON.stringify(newPeople))
/*
[{
  "firstName": "Jihad",
  "lastName": "Saladin",
  "addresses": [
    {
      "street": "12 Beaver Court",
      "city": "Snowmass",
      "state": "CO",
      "zip": "81615"
    },
    {
      "street": "16 Vail Rd",
      "city": "Vail",
      "state": "CO",
      "zip": "81657"
    }
  ]
}]
*/

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请不要删除问题的格式。您可以选择代码，单击Ctrl+k，编辑器将为您自动格式化tt（供下次参考）。如果您询问“更好的解决方案”，则您需要在某个地方有一个可比较的解决方案。通常你至少应该自己尝试一些东西。