LinkedList usage in C Sharp (C#) - Mini Project: Build & Apply

LinkedList usage

📖 Scenario: You are managing a waiting list for a small cafe. Customers arrive and are added to the list in order. Sometimes, customers leave the list before being served.

🎯 Goal: You will create a linked list to hold customer names, add customers to the list, remove a customer who leaves early, and finally display the current waiting list.

📋 What You'll Learn

Use the LinkedList<string> class

Add customers to the linked list using AddLast

Remove a specific customer using Remove

Iterate over the linked list to print all customer names

💡 Why This Matters

🌍 Real World

Linked lists are useful when you need to add or remove items quickly from the start or end of a list, like managing queues or waiting lists.

💼 Career

Understanding linked lists helps in software development roles that involve data structures, improving your ability to write efficient and flexible code.

1

Create the linked list

Create a LinkedList<string> called waitingList and add these customers in order using AddLast: "Alice", "Bob", "Charlie".

C Sharp (C#)

using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        // Create LinkedList waitingList and add "Alice", "Bob", "Charlie" using AddLast
        
    }
}

Hint

Use LinkedList<string> waitingList = new LinkedList<string>(); to create the list. Then add each name with waitingList.AddLast(name);.

2

Remove a customer

Create a variable called customerToRemove and set it to "Bob". Then remove customerToRemove from waitingList using the Remove method.

C Sharp (C#)

using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        LinkedList waitingList = new LinkedList();
        waitingList.AddLast("Alice");
        waitingList.AddLast("Bob");
        waitingList.AddLast("Charlie");
        // Create string customerToRemove = "Bob" and remove it from waitingList
        
    }
}

Hint

Set string customerToRemove = "Bob"; and then call waitingList.Remove(customerToRemove);.

3

Iterate over the linked list

Use a foreach loop with variable customer to go through waitingList and add each customer to a new List<string> called currentCustomers.

C Sharp (C#)

using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        LinkedList waitingList = new LinkedList();
        waitingList.AddLast("Alice");
        waitingList.AddLast("Bob");
        waitingList.AddLast("Charlie");
        string customerToRemove = "Bob";
        waitingList.Remove(customerToRemove);
        // Create List currentCustomers and add each customer from waitingList
        
    }
}

Hint

Create List<string> currentCustomers = new List<string>(); then use foreach (string customer in waitingList) to add each customer.

4

Print the current waiting list

Use a foreach loop with variable customer to print each name in currentCustomers on its own line using Console.WriteLine.

C Sharp (C#)

using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        LinkedList waitingList = new LinkedList();
        waitingList.AddLast("Alice");
        waitingList.AddLast("Bob");
        waitingList.AddLast("Charlie");
        string customerToRemove = "Bob";
        waitingList.Remove(customerToRemove);
        List currentCustomers = new List();
        foreach (string customer in waitingList)
        {
            currentCustomers.Add(customer);
        }
        // Print each customer in currentCustomers on its own line
        
    }
}

Hint

Use foreach (string customer in currentCustomers) and inside the loop write Console.WriteLine(customer);.

Practice

1. What is a key characteristic of a LinkedList in C#?

easy

A. It stores elements in nodes linked by references.

B. It stores elements in a fixed-size array.

C. It only allows adding elements at the end.

D. It cannot remove elements once added.

Solution

1. Step 1: Understand LinkedList structure

   A LinkedList stores elements in nodes, where each node points to the next (and possibly previous) node.

2. Step 2: Compare options with LinkedList behavior

   Only It stores elements in nodes linked by references. correctly describes this linked node structure; others describe arrays or incorrect behaviors.

3. Final Answer:

   It stores elements in nodes linked by references. -> Option A

4. Quick Check:

   LinkedList = nodes linked by references [OK]

Hint: LinkedList uses nodes connected by links, not arrays. [OK]

Common Mistakes:

• Thinking LinkedList uses arrays internally
• Assuming LinkedList only adds at the end
• Believing LinkedList cannot remove elements

2. Which of the following is the correct way to add an element at the start of a LinkedList<int> named list?

easy

A. list.AddStart(10);

B. list.AddFirst(10);

C. list.InsertAt(0, 10);

D. list.PushFront(10);

Solution

1. Step 1: Recall LinkedList method names

   The method to add an element at the start is AddFirst.

2. Step 2: Check each option's validity

   Only AddFirst is a valid LinkedList method; others are invalid or do not exist.

3. Final Answer:

   list.AddFirst(10); -> Option B

4. Quick Check:

   AddFirst adds at start [OK]

Hint: Use AddFirst to add at the start of LinkedList. [OK]

Common Mistakes:

• Using non-existent methods like AddStart or PushFront
• Confusing LinkedList with List methods
• Trying to use InsertAt which LinkedList does not have

3. What will be the output of this C# code?

var list = new LinkedList<string>();
list.AddLast("apple");
list.AddFirst("banana");
list.AddLast("cherry");
foreach(var item in list) Console.Write(item + " ");

medium

A. banana apple cherry

B. apple banana cherry

C. cherry apple banana

D. banana cherry apple

Solution

1. Step 1: Track insertion order

   First, "apple" is added last, so list: apple. Then "banana" added first, so list: banana, apple. Then "cherry" added last, so list: banana, apple, cherry.

2. Step 2: Understand foreach iteration order

   Foreach iterates from first to last node, so output is "banana apple cherry ".

3. Final Answer:

   banana apple cherry -> Option A

4. Quick Check:

   First = banana, last = cherry [OK]

Hint: AddFirst puts item at start; AddLast at end. [OK]

Common Mistakes:

• Assuming AddLast adds at start
• Confusing order of AddFirst and AddLast
• Expecting output in reverse order

4. Identify the error in this code snippet:

var list = new LinkedList<int>();
list.AddFirst(1);
list.AddLast(2);
list.Remove(3);
Console.WriteLine(list.Count);

medium

A. Remove(3) throws an exception because 3 is not in the list.

B. Count property does not exist on LinkedList.

C. AddFirst and AddLast methods are invalid for LinkedList.

D. Remove(3) does nothing since 3 is not found; Count remains 2.

Solution

1. Step 1: Understand Remove behavior

   Remove(value) tries to remove the first node with that value. If not found, it does nothing and returns false; no exception is thrown.

2. Step 2: Check Count after removal attempt

   Since 3 is not in the list, list remains with 2 elements; Count is 2.

3. Final Answer:

   Remove(3) does nothing since 3 is not found; Count remains 2. -> Option D

4. Quick Check:

   Remove missing value = no error, Count unchanged [OK]

Hint: Remove missing item does not throw error, just returns false. [OK]

Common Mistakes:

• Expecting Remove to throw exception if item missing
• Thinking AddFirst/AddLast are invalid
• Assuming Count is not a property

5. Given a LinkedList<int> with values 1, 2, 3, 4, 5, which code snippet correctly removes all even numbers from the list?

hard

A. foreach(var node in list) { if(node % 2 == 0) list.Remove(node); }

B. for(int i = 0; i < list.Count; i++) { if(list.ElementAt(i) % 2 == 0) list.Remove(list.ElementAt(i)); }

C. var current = list.First; while(current != null) { var next = current.Next; if(current.Value % 2 == 0) list.Remove(current); current = next; }

D. list.RemoveAll(x => x % 2 == 0);

Solution

1. Step 1: Understand safe removal during iteration

   Removing nodes while iterating requires storing next node before removal to avoid invalid references.

2. Step 2: Analyze each option

   foreach(var node in list) { if(node % 2 == 0) list.Remove(node); } uses foreach which throws error on modification during iteration. var current = list.First; while(current != null) { var next = current.Next; if(current.Value % 2 == 0) list.Remove(current); current = next; } correctly uses a while loop with next node saved. for(int i = 0; i < list.Count; i++) { if(list.ElementAt(i) % 2 == 0) list.Remove(list.ElementAt(i)); } uses ElementAt which is inefficient and unsafe. list.RemoveAll(x => x % 2 == 0); is invalid as LinkedList has no RemoveAll method.

3. Final Answer:

   var current = list.First; while(current != null) { var next = current.Next; if(current.Value % 2 == 0) list.Remove(current); current = next; } -> Option C

4. Quick Check:

   Use while loop with next saved to remove nodes safely [OK]

Hint: Save next node before removal to avoid iteration errors. [OK]

Common Mistakes:

• Modifying list inside foreach causes runtime error
• Using RemoveAll which LinkedList does not have
• Using ElementAt which is inefficient and unsafe