Removing the specified node from the LinkedList in C#?

The LinkedList<T> class in C# provides several methods to remove nodes from the linked list. You can remove nodes by value using Remove() method or remove specific node references using Remove(LinkedListNode<T>).

Syntax

Following are the different syntaxes for removing nodes from a LinkedList −

// Remove by value (removes first occurrence)
bool Remove(T value)

// Remove specific node
void Remove(LinkedListNode<T> node)

// Remove first node
void RemoveFirst()

// Remove last node
void RemoveLast()

Return Value

The Remove(T value) method returns true if the element is successfully removed; otherwise, false. Other removal methods are void and do not return values.

Using Remove() Method with Values

The Remove() method removes the first occurrence of the specified value from the LinkedList −

using System;
using System.Collections.Generic;

public class Demo {
   public static void Main() {
      LinkedList<int> list = new LinkedList<int>();
      list.AddLast(100);
      list.AddLast(200);
      list.AddLast(300);
      list.AddLast(400);
      list.AddLast(500);
      list.AddLast(300);
      list.AddLast(500);
      
      Console.WriteLine("LinkedList elements...");
      foreach(int i in list) {
         Console.WriteLine(i);
      }
      
      // Remove first occurrence of 500
      bool removed = list.Remove(500);
      Console.WriteLine("Removed 500: " + removed);
      
      Console.WriteLine("LinkedList elements...UPDATED");
      foreach(int i in list) {
         Console.WriteLine(i);
      }
   }
}

The output of the above code is −

LinkedList elements...
100
200
300
400
500
300
500
Removed 500: True
LinkedList elements...UPDATED
100
200
300
400
300
500

Using Remove() with String Values

The same approach works with string LinkedLists −

using System;
using System.Collections.Generic;

public class Demo {
   public static void Main() {
      LinkedList<string> list = new LinkedList<string>();
      list.AddLast("Mark");
      list.AddLast("David");
      list.AddLast("Harry");
      list.AddLast("John");
      list.AddLast("Kevin");
      
      Console.WriteLine("LinkedList elements...");
      foreach(string str in list) {
         Console.WriteLine(str);
      }
      
      // Remove Harry from the list
      bool removed = list.Remove("Harry");
      Console.WriteLine("Removed Harry: " + removed);
      
      Console.WriteLine("LinkedList elements...UPDATED");
      foreach(string str in list) {
         Console.WriteLine(str);
      }
   }
}

The output of the above code is −

LinkedList elements...
Mark
David
Harry
John
Kevin
Removed Harry: True
LinkedList elements...UPDATED
Mark
David
John
Kevin

Removing Specific Node References

You can also remove specific node references found using Find() or FindLast() methods −

using System;
using System.Collections.Generic;

public class Demo {
   public static void Main() {
      LinkedList<int> list = new LinkedList<int>();
      list.AddLast(10);
      list.AddLast(20);
      list.AddLast(30);
      list.AddLast(20);
      list.AddLast(40);
      
      Console.WriteLine("Original LinkedList:");
      foreach(int i in list) {
         Console.WriteLine(i);
      }
      
      // Find and remove the last occurrence of 20
      LinkedListNode<int> node = list.FindLast(20);
      if(node != null) {
         list.Remove(node);
         Console.WriteLine("Removed last occurrence of 20");
      }
      
      Console.WriteLine("Updated LinkedList:");
      foreach(int i in list) {
         Console.WriteLine(i);
      }
   }
}

The output of the above code is −

Original LinkedList:
10
20
30
20
40
Removed last occurrence of 20
Updated LinkedList:
10
20
30
40

Comparison of Removal Methods

Conclusion

The LinkedList<T> class provides flexible methods for removing nodes. Use Remove(T value) to remove by value (first occurrence only), or use Remove(LinkedListNode<T>) with Find() or FindLast() to remove specific node occurrences. The Remove(T value) method returns a boolean indicating success or failure.