Question

我在面试中解决了问题。他们提出了一个json对象：

putchar(A)

需要组织结构的格式输出，如：

{
    "UserName": "Tom Las",
    "title": "Director"
},
{
    "UserName": "Mike Sea",
    "title": "senior manager"
},
{
    "UserName": "Jojo Lee",
    "title": "manager",
    "direct Report": "Mike Sea"
},
{
    "UserName": "Luke Shi",
    "title": "manager",
    "direct Report": "Mike Sea"
},
{
    "UserName": "Bob Aeo",
    "title": "engineer",
    "direct Report": "Luke Shi"
},
{
    "UserName": "Zobu hu",
    "title": "engineer",
    "direct Report": "Tom Las"
}

这是我的解决方案。我使用两个字典来跟踪领导者和团队成员信息并使用＆＃34; ＆＃34; * n缩进子团队中的团队成员。我觉得下面的代码是笨拙的，并且不能灵活地显示更多级别的子团队，例如，如果＆＃34; Bob Aeo＆＃34;他下面也有团队成员。在这种情况下，我应该考虑不同的数据结构而不是字典吗？谢谢！

Mike Sea - senior manager
  Jojo Lee
  Luke Shi
    Bob Aeo
Tom Las - Director
  Zobu hu

Answer 1

建立一个树...我将“直接报告”更改为“经理”，因为我认为重点是在MemberInfo对象中命名该人的经理。管理器的反向链接是直接报告列表，因此我在下面的示例中设置了该列表。我开始索引列表，添加一个"direct reports"列表，然后填写该列表。然后，这只是一个递归例程的问题，打印和输入后跟任何直接报告。代码没有假设组织是多么臃肿，但是假设列表中没有循环......我曾经在自己的工作中向自己报告并且对各种软件造成了严重破坏。有趣的是，我当时正致力于管理Active Directory Manager和directReports字段的代码。

MemberInfo = [
    {
        "UserName": "Tom Las",
        "title": "Director"
    },
    {
        "UserName": "Mike Sea",
        "title": "senior manager"
    },
    {
        "UserName": "Jojo Lee",
        "title": "manager",
        "Manager": "Mike Sea"
    },
    {
        "UserName": "Luke Shi",
        "title": "manager",
        "Manager": "Mike Sea"
    },
    {
        "UserName": "Bob Aeo",
        "title": "engineer",
        "Manager": "Luke Shi"
    },
    {
        "UserName": "Zobu hu",
        "title": "engineer",
        "Manager": "Tom Las"
    }
]

def crawl_ranks(name, indent=''):
    """Prints member and direct report names recursively"""
    member = member_index[name]
    print('{}{}'.format(indent, member['UserName']))
    for name in sorted(member.get('direct reports', [])):
        crawl_ranks(name, indent=indent + '  ')

# index by name for easy lookup
member_index = {member['UserName']:member for member in MemberInfo}

# add direct reports list to member info
for member in MemberInfo:
    member['direct reports'] = []

# add dummy index entry for bossless persons
member_index[None] = {'direct reports':[]}

# add member name to boss's direct reports list
for member in MemberInfo:
    member_index[member.get('Manager')]['direct reports'].append(member['UserName'])

# print, starting with top-level managers
for name in sorted(member_index[None]['direct reports']):
    crawl_ranks(name)

Python优化数据结构

1 个答案: