Sort a Linked List Using Merge Sort in DSA C# - Build from Scratch

Sort a Linked List Using Merge Sort

📖 Scenario: You are working on a contact management app. The contacts are stored in a linked list by the order they were added. Now, you want to sort the contacts alphabetically by their names to make searching easier.

🎯 Goal: Build a program that creates a linked list of contacts, then sorts this linked list alphabetically using the merge sort algorithm designed for linked lists.

📋 What You'll Learn

Create a linked list with nodes containing contact names as strings.

Implement a method to split the linked list into two halves.

Implement the merge sort algorithm to sort the linked list alphabetically.

Print the sorted linked list showing the contact names in order.

💡 Why This Matters

🌍 Real World

Sorting linked lists is useful in applications like contact managers, music playlists, or any system where data is stored in linked lists and needs to be ordered efficiently.

💼 Career

Understanding linked list sorting algorithms like merge sort is important for software developers working with data structures, improving performance and memory usage in real-world applications.

1

Create the Linked List with Contact Names

Create a class called Node with a string property Name and a Node property Next. Then create a linked list with these exact contact names in order: "John", "Alice", "Bob", "Diana", "Charlie". Store the head of the list in a variable called head.

DSA C#

// Define Node class and create linked list with given names
// Your code here

Hint

Define the Node class first. Then link nodes by setting the Next property to the next node.

2

Add a Helper Method to Split the Linked List

Add a method called GetMiddle that takes a Node parameter called head and returns the middle node of the linked list using the slow and fast pointer technique.

DSA C#

public class Node
{
    public string Name;
    public Node Next;
    public Node(string name)
    {
        Name = name;
        Next = null;
    }
}

Node head = new Node("John");
head.Next = new Node("Alice");
head.Next.Next = new Node("Bob");
head.Next.Next.Next = new Node("Diana");
head.Next.Next.Next.Next = new Node("Charlie");

// Add GetMiddle method below
// Your code here

Hint

Use two pointers: slow moves one step, fast moves two steps until fast reaches the end.

3

Implement Merge Sort for the Linked List

Add a method called MergeSort that takes a Node parameter head and returns the head of the sorted linked list. Use the GetMiddle method to split the list, recursively sort each half, and merge them using a helper method SortedMerge. Also add the SortedMerge method that merges two sorted linked lists alphabetically by Name.

DSA C#

public class Node
{
    public string Name;
    public Node Next;
    public Node(string name)
    {
        Name = name;
        Next = null;
    }
}

Node head = new Node("John");
head.Next = new Node("Alice");
head.Next.Next = new Node("Bob");
head.Next.Next.Next = new Node("Diana");
head.Next.Next.Next.Next = new Node("Charlie");

Node GetMiddle(Node head)
{
    if (head == null) return head;
    Node slow = head;
    Node fast = head.Next;
    while (fast != null && fast.Next != null)
    {
        slow = slow.Next;
        fast = fast.Next.Next;
    }
    return slow;
}

// Add MergeSort and SortedMerge methods below
// Your code here

Hint

Use recursion to split and sort each half, then merge sorted halves by comparing names alphabetically.

4

Print the Sorted Linked List

Call the MergeSort method on the head variable and store the result back in head. Then print the sorted linked list by traversing from head and printing each Name separated by -> . End with null.

DSA C#

public class Node
{
    public string Name;
    public Node Next;
    public Node(string name)
    {
        Name = name;
        Next = null;
    }
}

Node head = new Node("John");
head.Next = new Node("Alice");
head.Next.Next = new Node("Bob");
head.Next.Next.Next = new Node("Diana");
head.Next.Next.Next.Next = new Node("Charlie");

Node GetMiddle(Node head)
{
    if (head == null) return head;
    Node slow = head;
    Node fast = head.Next;
    while (fast != null && fast.Next != null)
    {
        slow = slow.Next;
        fast = fast.Next.Next;
    }
    return slow;
}

Node SortedMerge(Node a, Node b)
{
    if (a == null) return b;
    if (b == null) return a;

    Node result;
    if (string.Compare(a.Name, b.Name) <= 0)
    {
        result = a;
        result.Next = SortedMerge(a.Next, b);
    }
    else
    {
        result = b;
        result.Next = SortedMerge(a, b.Next);
    }
    return result;
}

Node MergeSort(Node head)
{
    if (head == null || head.Next == null)
        return head;

    Node middle = GetMiddle(head);
    Node nextOfMiddle = middle.Next;
    middle.Next = null;

    Node left = MergeSort(head);
    Node right = MergeSort(nextOfMiddle);

    Node sortedList = SortedMerge(left, right);
    return sortedList;
}

// Sort the list and print it
// Your code here

Hint

After sorting, print each node's Name followed by -> , ending with null.

Practice

1. What is the main advantage of using merge sort to sort a linked list compared to other sorting algorithms?

easy

A. It efficiently sorts the list in O(n log n) time without needing extra space for arrays.

B. It sorts the list in O(n^2) time but uses less memory.

C. It requires the list to be converted into an array first.

D. It only works for linked lists with integer values.

Solution

1. Step 1: Understand merge sort time complexity on linked lists

   Merge sort divides the list into halves recursively and merges them, achieving O(n log n) time.

2. Step 2: Compare space usage with other sorts

   Unlike quicksort or heapsort on arrays, merge sort on linked lists does not require extra array space, making it memory efficient.

3. Final Answer:

   It efficiently sorts the list in O(n log n) time without needing extra space for arrays. -> Option A

4. Quick Check:

   Merge sort = O(n log n) + memory efficient [OK]

Hint: Merge sort is best for linked lists due to O(n log n) and no extra arrays [OK]

Common Mistakes:

• Thinking merge sort uses extra array space for linked lists
• Confusing merge sort with bubble sort time complexity
• Assuming merge sort only works on arrays
• Believing merge sort requires integer-only lists

2. Which of the following is the correct way to split a linked list into two halves in C# for merge sort?

easy

A. Convert the list to an array and split the array.

B. Use a single pointer to count nodes and split at half index.

C. Use two pointers: slow moves 1 step, fast moves 2 steps; slow marks middle.

D. Split the list by removing the last node repeatedly.

Solution

1. Step 1: Identify the standard technique to find middle in linked list

   Using two pointers, slow and fast, where fast moves twice as fast, slow ends at middle.

2. Step 2: Explain why other options are incorrect

   Counting nodes requires extra traversal; converting to array is inefficient; removing last node is not splitting in half.

3. Final Answer:

   Use two pointers: slow moves 1 step, fast moves 2 steps; slow marks middle. -> Option C

4. Quick Check:

   Two-pointer technique = correct split [OK]

Hint: Use slow-fast pointers to find middle in one pass [OK]

Common Mistakes:

• Trying to split by counting nodes first
• Converting linked list to array unnecessarily
• Removing nodes instead of splitting
• Using only one pointer to find middle

3. Given the linked list: 4 -> 2 -> 1 -> 3 -> null, what is the sorted list after applying merge sort?

medium

A. 4 -> 3 -> 2 -> 1 -> null

B. 1 -> 3 -> 2 -> 4 -> null

C. 2 -> 1 -> 3 -> 4 -> null

D. 1 -> 2 -> 3 -> 4 -> null

Solution

1. Step 1: Break down the list and sort halves recursively

   Split into (4 -> 2) and (1 -> 3). Sort each half: (2 -> 4) and (1 -> 3).

2. Step 2: Merge the two sorted halves

   Merge (2 -> 4) and (1 -> 3) to get (1 -> 2 -> 3 -> 4).

3. Final Answer:

   1 -> 2 -> 3 -> 4 -> null -> Option D

4. Quick Check:

   Sorted ascending = 1,2,3,4 [OK]

Hint: Merge sorted halves step-by-step to get final sorted list [OK]

Common Mistakes:

• Not sorting halves before merging
• Merging without sorting causing wrong order
• Confusing node values during merge
• Forgetting to set last node's next to null

4. In the following C# merge sort code for linked list, what is the bug?

public ListNode MergeSort(ListNode head) {
  if (head == null || head.next == null) return head;
  ListNode mid = GetMiddle(head);
  ListNode left = MergeSort(head);
  ListNode right = MergeSort(mid);
  return Merge(left, right);
}

medium

A. The base case is incorrect; it should check for head.next.next.

B. The middle node is not disconnected from the left half, causing infinite recursion.

C. The Merge function is called before sorting the halves.

D. The GetMiddle function returns the last node instead of the middle.

Solution

1. Step 1: Identify the role of splitting the list

   After finding middle, the list must be split by setting middle's previous node next to null to separate halves.

2. Step 2: Explain the bug effect

   Without disconnecting, recursive calls keep processing the same list, causing infinite recursion or stack overflow.

3. Final Answer:

   The middle node is not disconnected from the left half, causing infinite recursion. -> Option B

4. Quick Check:

   Disconnect middle to avoid infinite recursion [OK]

Hint: Always break list into two by setting middle's previous next to null [OK]

Common Mistakes:

• Not disconnecting the list at middle
• Wrong base case conditions
• Calling merge before sorting halves
• Assuming GetMiddle returns last node

5. You have a linked list with nodes containing duplicate values: 3 -> 1 -> 2 -> 3 -> 2 -> null. After applying merge sort, what will be the correct sorted linked list?

hard

A. 1 -> 2 -> 2 -> 3 -> 3 -> null

B. 3 -> 3 -> 2 -> 2 -> 1 -> null

C. 1 -> 3 -> 2 -> 3 -> 2 -> null

D. 2 -> 1 -> 3 -> 2 -> 3 -> null

Solution

1. Step 1: Understand merge sort handles duplicates by sorting all nodes

   Merge sort compares node values and merges in ascending order, preserving duplicates in sorted order.

2. Step 2: Sort the list step-by-step

   Sorted order is 1, 2, 2, 3, 3, so the final list is 1 -> 2 -> 2 -> 3 -> 3 -> null.

3. Final Answer:

   1 -> 2 -> 2 -> 3 -> 3 -> null -> Option A

4. Quick Check:

   Duplicates sorted ascending = 1,2,2,3,3 [OK]

Hint: Merge sort keeps duplicates in sorted order naturally [OK]

Common Mistakes:

• Assuming duplicates get removed
• Mixing order of duplicate nodes
• Not sorting duplicates correctly
• Confusing stable sort with unstable sort