如何在自引用linq中获得针对高度父节点的极值叶节点
如何在自引用linq中获得针对高度父节点的极值叶节点
提问于 2017-11-03 14:37:44
我有了数据结构
-Emp1 - Emp1_1 - Emp1_1_1
- Emp1_1_2
- Emp1_2 - Emp1_2_1
- Emp1_2_2
-Emp2 - Emp2_1 - Emp2_1_1
- Emp2_1_2
- Emp2_2 - Emp2_2_1
- Emp2_2_2我想要一个字典，它拥有极端的叶节点作为关键字和它的最高父作为值。操作：
(Emp1_1_1，Emp1) (Emp1_1_2，Emp1) (Emp1_2_1，Emp1) (Emp1_2_2，Emp1) (Emp2_1_1，Emp2) (Emp2_1_2，Emp2) (Emp2_2_1，Emp2) (Emp2_2_2，Emp2)
(Emp1_1_1, Emp1)
(Emp1_1_2, Emp1)
(Emp1_2_1, Emp1)
(Emp1_2_2, Emp1)
(Emp2_1_1, Emp2)
(Emp2_1_2, Emp2)
(Emp2_2_1, Emp2)
(Emp2_2_2, Emp2)下面是c#代码
public class Program
{
public static void Main()
{
var program = new Program();
var empList = program.GetData();
//Some linq which return the expected dictionary.
Console.WriteLine("Press any key to exit");
Console.ReadKey();
}
public class Employee
{
public int Id { get; set; }
public string Name { get; set; }
public int parentEmployee { get; set; }
public List<Employee> childList { get; set; }
}
List<Employee> GetData()
{
var empList = new List<Employee>();
var emp1 = new Employee() { Id = 1, Name = "Emp1_Name", childList = new List<Employee>() };
var emp1_1 = new Employee() { Id = 2, Name = "Emp1_1_Name", parentEmployee = emp1.Id, childList = new List<Employee>() };
var emp1_2 = new Employee() { Id = 3, Name = "Emp1_2_Name", parentEmployee = emp1.Id, childList = new List<Employee>() };
emp1.childList.Add(emp1_1);
emp1.childList.Add(emp1_2);
var emp1_1_1 = new Employee() { Id = 4, Name = "Emp1_1_1_Name", parentEmployee = emp1_1.Id };
var emp1_1_2 = new Employee() { Id = 5, Name = "Emp1_1_2_Name", parentEmployee = emp1_1.Id };
emp1_1.childList.Add(emp1_1_1);
emp1_1.childList.Add(emp1_1_2);
var emp1_2_1 = new Employee() { Id = 6, Name = "Emp1_2_1_Name", parentEmployee = emp1_2.Id };
var emp1_2_2 = new Employee() { Id = 7, Name = "Emp1_2_2_Name", parentEmployee = emp1_2.Id };
emp1_2.childList.Add(emp1_2_1);
emp1_2.childList.Add(emp1_2_2);
var emp2 = new Employee() { Id = 1, Name = "emp2_Name", childList = new List<Employee>() };
var emp2_1 = new Employee() { Id = 2, Name = "emp2_1_Name", parentEmployee = emp2.Id, childList = new List<Employee>() };
var emp2_2 = new Employee() { Id = 3, Name = "emp2_2_Name", parentEmployee = emp2.Id, childList = new List<Employee>() };
emp2.childList.Add(emp2_1);
emp2.childList.Add(emp2_2);
var emp2_1_1 = new Employee() { Id = 4, Name = "emp2_1_1_Name", parentEmployee = emp2_1.Id };
var emp2_1_2 = new Employee() { Id = 5, Name = "emp2_1_2_Name", parentEmployee = emp2_1.Id };
emp2_1.childList.Add(emp2_1_1);
emp2_1.childList.Add(emp2_1_2);
var emp2_2_1 = new Employee() { Id = 6, Name = "emp2_2_1_Name", parentEmployee = emp2_2.Id };
var emp2_2_2 = new Employee() { Id = 7, Name = "emp2_2_2_Name", parentEmployee = emp2_2.Id };
emp2_2.childList.Add(emp2_2_1);
emp2_2.childList.Add(emp2_2_2);
empList.Add(emp1);
empList.Add(emp2);
return empList;
}
}我尝试了一些解决方案，比如在linq中调用私有自引用函数，但它适用于一父一子关系。
回答 2
Stack Overflow用户
发布于 2017-11-03 16:03:24
如果可以使用递归函数：
Func<IEnumerable<Employee>, IEnumerable<Employee>> flatten = null;
flatten = (IEnumerable<Employee> employees) =>
employees.SelectMany(c => c.childList != null ? flatten(c.childList) : Enumerable.Empty<Employee>()).Concat(employees);
var dict = (
from topNode in empList
from node in flatten(topNode.childList)
where node.childList == null || node.childList.Count == 0
select new KeyValuePair<string, string>(node.Name, topNode.Name)
).ToDictionary(kv => kv.Key, kv => kv.Value);
foreach(var keyValuePair in dict)
{
Console.WriteLine($"Key={keyValuePair.Key}, Value={keyValuePair.Value}");
}这将输出以下内容：
Key=Emp1_1_1_Name, Value=Emp1_Name
Key=Emp1_1_2_Name, Value=Emp1_Name
Key=Emp1_2_1_Name, Value=Emp1_Name
Key=Emp1_2_2_Name, Value=Emp1_Name
Key=emp2_1_1_Name, Value=emp2_Name
Key=emp2_1_2_Name, Value=emp2_Name
Key=emp2_2_1_Name, Value=emp2_Name
Key=emp2_2_2_Name, Value=emp2_NameStack Overflow用户
发布于 2017-11-03 18:11:57
尝尝这个
public static void Main()
{
var program = new Program();
var empList = program.GetData();
Func<IEnumerable<Employee>, Employee, Dictionary<Employee, Employee>> TreeToDict = null;
TreeToDict = (employees, topParent) =>
employees.SelectMany(e =>
{
if(e.childList != null)
{
return TreeToDict(e.childList, topParent == null ? e : topParent);
}
else
{
Dictionary<Employee, Employee> dict = new Dictionary<Employee, Employee>();
dict.Add(e, topParent == null ? e : topParent);
return dict;
}
}).ToDictionary(x => x.Key, x => x.Value);
foreach(var emp in TreeToDict (empList, null))
{
Console.WriteLine("key : {0}, value {1}", emp.Key.Name, emp.Value.Name);
}
}结果
key : Emp1_1_1_Name, value Emp1_Name
key : Emp1_1_2_Name, value Emp1_Name
key : Emp1_2_1_Name, value Emp1_Name
key : Emp1_2_2_Name, value Emp1_Name
key : emp2_1_1_Name, value emp2_Name
key : emp2_1_2_Name, value emp2_Name
key : emp2_2_1_Name, value emp2_Name
key : emp2_2_2_Name, value emp2_Name页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/47089835
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