Merge Two Sorted Linked Lists in DSA C++ - Build from Scratch

Merge Two Sorted Linked Lists

📖 Scenario: You are working on a contact list app. You have two separate lists of contacts sorted by their ID numbers. You want to combine these two lists into one sorted list so that users can see all contacts in order.

🎯 Goal: Build a program that merges two sorted singly linked lists into one sorted linked list.

📋 What You'll Learn

Create two sorted singly linked lists with given values

Use a pointer to traverse and merge the lists

Create a new sorted linked list by merging the two lists

Print the merged linked list in order

💡 Why This Matters

🌍 Real World

Merging sorted lists is common in apps like contact managers, calendars, or any system that combines sorted data streams.

💼 Career

Understanding linked list merging is important for software engineers working on data processing, system design, and algorithm optimization.

1

Create Two Sorted Linked Lists

Create a singly linked list node struct called ListNode with an int val and a ListNode* next. Then create two sorted linked lists called list1 and list2 with these exact values: list1 contains 1 -> 3 -> 5 -> null, and list2 contains 2 -> 4 -> 6 -> null.

DSA C++

// Define ListNode struct
// Create list1 with nodes 1 -> 3 -> 5
// Create list2 with nodes 2 -> 4 -> 6

Hint

Define the node structure first. Then link nodes by setting the next pointer to the next node.

2

Create a Dummy Node and a Pointer for Merging

Create a dummy ListNode called dummy with value 0. Create a pointer current and set it to point to dummy. This will help build the merged list.

DSA C++

#include 

struct ListNode {
    int val;
    ListNode* next;
    ListNode(int x) : val(x), next(nullptr) {}
};

// Create list1: 1 -> 3 -> 5
ListNode* list1 = new ListNode(1);
list1->next = new ListNode(3);
list1->next->next = new ListNode(5);

// Create list2: 2 -> 4 -> 6
ListNode* list2 = new ListNode(2);
list2->next = new ListNode(4);
list2->next->next = new ListNode(6);

// Create dummy node and current pointer
// Your code here

Hint

The dummy node helps simplify edge cases. The current pointer will move along the new list as you add nodes.

3

Merge the Two Sorted Lists

Use a while loop to compare list1->val and list2->val. Attach the smaller node to current->next and move that list's pointer forward. Move current forward as well. After the loop, attach the remaining nodes from list1 or list2 to current->next.

DSA C++

#include 

struct ListNode {
    int val;
    ListNode* next;
    ListNode(int x) : val(x), next(nullptr) {}
};

// Create list1: 1 -> 3 -> 5
ListNode* list1 = new ListNode(1);
list1->next = new ListNode(3);
list1->next->next = new ListNode(5);

// Create list2: 2 -> 4 -> 6
ListNode* list2 = new ListNode(2);
list2->next = new ListNode(4);
list2->next->next = new ListNode(6);

// Create dummy node and current pointer
ListNode* dummy = new ListNode(0);
ListNode* current = dummy;

// Merge lists
// Your code here

Hint

Keep moving the current pointer and attach the smaller node each time. After the loop, attach the leftover nodes.

4

Print the Merged Linked List

Use a ListNode* pointer called temp to traverse the merged list starting from dummy->next. Print each node's val followed by -> . After the last node, print null.

DSA C++

#include 

struct ListNode {
    int val;
    ListNode* next;
    ListNode(int x) : val(x), next(nullptr) {}
};

// Create list1: 1 -> 3 -> 5
ListNode* list1 = new ListNode(1);
list1->next = new ListNode(3);
list1->next->next = new ListNode(5);

// Create list2: 2 -> 4 -> 6
ListNode* list2 = new ListNode(2);
list2->next = new ListNode(4);
list2->next->next = new ListNode(6);

// Create dummy node and current pointer
ListNode* dummy = new ListNode(0);
ListNode* current = dummy;

while (list1 != nullptr && list2 != nullptr) {
    if (list1->val val) {
        current->next = list1;
        list1 = list1->next;
    } else {
        current->next = list2;
        list2 = list2->next;
    }
    current = current->next;
}

if (list1 != nullptr) {
    current->next = list1;
} else {
    current->next = list2;
}

// Print merged list
// Your code here

Hint

Start from dummy->next because dummy is a placeholder. Print each value followed by -> . End with null.

Practice

1. What is the main goal when merging two sorted linked lists into one sorted linked list?

easy

A. Sort each list separately before merging

B. Reverse the order of nodes in both lists

C. Combine nodes from both lists maintaining sorted order

D. Remove duplicate nodes from both lists

Solution

1. Step 1: Understand the merge operation

   Merging means combining two lists into one while keeping the order sorted.

2. Step 2: Maintain sorted order during merge

   We pick the smaller node from the heads of both lists repeatedly to keep the merged list sorted.

3. Final Answer:

   Combine nodes from both lists maintaining sorted order -> Option C

4. Quick Check:

   Merge = Combine sorted lists [OK]

Hint: Always pick the smaller node first to keep order [OK]

Common Mistakes:

• Trying to reverse lists instead of merging
• Sorting lists again after merging
• Ignoring order and just concatenating

2. Which of the following is the correct way to declare a node structure for a singly linked list in C++?

easy

A. struct Node { int data; Node next; };

B. class Node { int data; Node next; };

C. class Node { int data; Node* next; };

D. struct Node { int data; Node* next; };

Solution

1. Step 1: Understand node structure in C++

   A node should have data and a pointer to the next node.

2. Step 2: Check pointer usage

   Using Node* next is correct to point to the next node; Node next would cause infinite size error.

3. Final Answer:

   struct Node { int data; Node* next; }; -> Option D

4. Quick Check:

   Node pointer = Node* next [OK]

Hint: Use pointers for next node in linked list [OK]

Common Mistakes:

• Using object instead of pointer for next node
• Missing semicolon after struct
• Using class without access specifier

3. What will be the output linked list after merging these two sorted lists?
List1: 1 -> 3 -> 5 -> null
List2: 2 -> 4 -> 6 -> null
Using a standard merge algorithm?

medium

A. 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null

B. 1 -> 3 -> 5 -> 2 -> 4 -> 6 -> null

C. 2 -> 4 -> 6 -> 1 -> 3 -> 5 -> null

D. 1 -> 2 -> 4 -> 3 -> 5 -> 6 -> null

Solution

1. Step 1: Compare heads of both lists

   Start with 1 (List1) and 2 (List2). 1 is smaller, so first node is 1.

2. Step 2: Continue picking smaller nodes

   Next compare 3 and 2, pick 2; then 3 and 4, pick 3; then 5 and 4, pick 4; then 5 and 6, pick 5; finally 6.

3. Final Answer:

   1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null -> Option A

4. Quick Check:

   Sorted merge = 1,2,3,4,5,6 [OK]

Hint: Pick smaller head node repeatedly [OK]

Common Mistakes:

• Appending one list after the other without sorting
• Mixing node order incorrectly
• Skipping nodes during merge

4. Given the following code snippet to merge two sorted linked lists, what is the error?

Node* merge(Node* l1, Node* l2) {
  if (!l1) return l2;
  if (!l2) return l1;
  if (l1->data < l2->data) {
    l1->next = merge(l1->next, l2);
    return l1;
  } else {
    l2->next = merge(l1, l2->next);
    return l2;
  }
}

medium

A. No error, code correctly merges two sorted lists

B. Missing base case for empty lists

C. Incorrect comparison operator used

D. Function does not handle cyclic lists

Solution

1. Step 1: Check base cases

   Code returns the other list if one is empty, which is correct.

2. Step 2: Check recursive merge logic

   It compares data correctly and recursively merges the rest, returning the smaller node.

3. Final Answer:

   No error, code correctly merges two sorted lists -> Option A

4. Quick Check:

   Recursive merge logic correct [OK]

Hint: Check base cases and recursion carefully [OK]

Common Mistakes:

• Forgetting to return merged list
• Wrong base case conditions
• Using wrong pointers in recursion

5. You have two sorted linked lists:
List1: 1 -> 4 -> 5 -> 7 -> null
List2: 2 -> 3 -> 6 -> 8 -> 9 -> null
Write the merged list after merging them. Then, if you want to remove duplicates from the merged list, what will be the final list?
Assume duplicates can appear after merge.

hard

A. Merged: 1 -> 4 -> 5 -> 7 -> 2 -> 3 -> 6 -> 8 -> 9 -> null; After removing duplicates: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null

B. Merged: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null; After removing duplicates: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null

C. Merged: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null; After removing duplicates: 1 -> 2 -> 3 -> 4 -> 5 -> null

D. Merged: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null; After removing duplicates: 1 -> 4 -> 5 -> 7 -> 2 -> 3 -> 6 -> 8 -> 9 -> null

Solution

1. Step 1: Merge both sorted lists

   Pick nodes in order: 1,2,3,4,5,6,7,8,9 forming a sorted merged list.

2. Step 2: Remove duplicates if any

   Since all values are unique, the list remains the same after removing duplicates.

3. Final Answer:

   Merged: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null; After removing duplicates: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null -> Option B

4. Quick Check:

   Sorted merge with unique values [OK]

Hint: Merge sorted, then remove duplicates if any [OK]

Common Mistakes:

• Concatenating lists without sorting
• Assuming duplicates always exist
• Removing nodes incorrectly