Merge Two Sorted Linked Lists in DSA Java - Build from Scratch

Merge Two Sorted Linked Lists

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

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

📋 What You'll Learn

Create two sorted linked lists using the ListNode class

Create a helper variable to start the merged list

Write the core logic to merge the two sorted lists into one sorted list

Print the merged linked list in the format: value -> value -> ... -> null

💡 Why This Matters

🌍 Real World

Merging sorted lists is common in contact apps, calendars, and any place where sorted data from different sources needs to be combined.

💼 Career

Understanding linked list merging is important for coding interviews and working with data structures in software development.

1

Create two sorted linked lists

Create a class called ListNode with an integer val and a next pointer. Then create two sorted linked lists called list1 and list2 with these exact values:
list1: 1 -> 3 -> 5 -> null
list2: 2 -> 4 -> 6 -> null

DSA Java

// Define ListNode class and create list1 and list2 with given values
// Your code here

Hint

Start by defining the ListNode class with a constructor. Then create the two lists by linking nodes with the given values.

2

Create a dummy node to start the merged list

Inside the main method, create a ListNode variable called dummy with value 0. Also create a ListNode variable called current and set it equal to dummy. This will help build the merged list.

DSA Java

class ListNode {
    int val;
    ListNode next;
    ListNode(int val) {
        this.val = val;
        this.next = null;
    }
}

public class Main {
    public static void main(String[] args) {
        ListNode list1 = new ListNode(1);
        list1.next = new ListNode(3);
        list1.next.next = new ListNode(5);

        ListNode list2 = new ListNode(2);
        list2.next = new ListNode(4);
        list2.next.next = new ListNode(6);

        // Create dummy and current nodes
        // Your code here
    }
}

Hint

The dummy node helps to easily return the head of the merged list later. The current pointer will move as you add nodes.

3

Merge the two sorted linked lists

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

DSA Java

class ListNode {
    int val;
    ListNode next;
    ListNode(int val) {
        this.val = val;
        this.next = null;
    }
}

public class Main {
    public static void main(String[] args) {
        ListNode list1 = new ListNode(1);
        list1.next = new ListNode(3);
        list1.next.next = new ListNode(5);

        ListNode list2 = new ListNode(2);
        list2.next = new ListNode(4);
        list2.next.next = new ListNode(6);

        ListNode dummy = new ListNode(0);
        ListNode current = dummy;

        // Merge lists
        // Your code here
    }
}

Hint

Use a loop to pick the smaller node from the two lists and attach it to the merged list. After the loop, attach the leftover nodes.

4

Print the merged linked list

Create a ListNode variable called merged and set it to dummy.next. Use a while loop to print each node's val followed by " -> ". After the loop, print "null" to show the end of the list.

DSA Java

class ListNode {
    int val;
    ListNode next;
    ListNode(int val) {
        this.val = val;
        this.next = null;
    }
}

public class Main {
    public static void main(String[] args) {
        ListNode list1 = new ListNode(1);
        list1.next = new ListNode(3);
        list1.next.next = new ListNode(5);

        ListNode list2 = new ListNode(2);
        list2.next = new ListNode(4);
        list2.next.next = new ListNode(6);

        ListNode dummy = new ListNode(0);
        ListNode current = dummy;

        while (list1 != null && list2 != null) {
            if (list1.val < list2.val) {
                current.next = list1;
                list1 = list1.next;
            } else {
                current.next = list2;
                list2 = list2.next;
            }
            current = current.next;
        }

        if (list1 != null) {
            current.next = list1;
        }

        if (list2 != null) {
            current.next = list2;
        }

        // Print merged list
        // Your code here
    }
}

Hint

Start from dummy.next and print each node's value followed by " -> ". End with printing "null".

Practice

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

easy

A. To create a new sorted linked list containing all elements from both lists

B. To reverse both linked lists

C. To find the maximum element in both lists

D. To delete duplicate elements from both lists

Solution

1. Step 1: Understand the input lists

   Two linked lists are sorted individually, meaning their elements are in ascending order.

2. Step 2: Merge while maintaining order

   The goal is to combine these two lists into one sorted list that contains all elements from both.

3. Final Answer:

   To create a new sorted linked list containing all elements from both lists -> Option A

4. Quick Check:

   Merging sorted lists means combining all elements in order [OK]

Hint: Merging means combining sorted lists into one sorted list [OK]

Common Mistakes:

• Thinking merge means reversing lists
• Assuming merge removes duplicates
• Confusing merge with finding max element

2. Which of the following Java method signatures correctly represents a function to merge two sorted linked lists of integers?

easy

A. public int merge(ListNode l1, ListNode l2)

B. public void merge(int l1, int l2)

C. public ListNode merge(ListNode l1, ListNode l2)

D. public ListNode merge(int[] l1, int[] l2)

Solution

1. Step 1: Identify correct parameter types

   The function should accept two linked lists, so parameters must be ListNode types, not primitives or arrays.

2. Step 2: Identify correct return type

   The function returns a merged linked list, so return type should be ListNode, not void or int.

3. Final Answer:

   public ListNode merge(ListNode l1, ListNode l2) -> Option C

4. Quick Check:

   Method with ListNode params and return type is correct [OK]

Hint: Method merges two ListNode inputs and returns ListNode [OK]

Common Mistakes:

• Using primitive types instead of ListNode
• Returning void instead of ListNode
• Using arrays instead of linked lists

3. What is the output linked list after merging these two sorted lists?

List 1: 1 -> 3 -> 5 -> null
List 2: 2 -> 4 -> 6 -> null

medium

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

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

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

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

Solution

1. Step 1: Compare first nodes of both lists

   1 (List 1) is less than 2 (List 2), so 1 is first in merged list.

2. Step 2: Continue comparing next nodes

   Next compare 3 and 2, 2 is smaller, add 2, then 3 vs 4, add 3, then 5 vs 4, add 4, then 5 vs 6, add 5, finally add 6.

3. Final Answer:

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

4. Quick Check:

   Merge keeps ascending order combining both lists [OK]

Hint: Pick smaller head node repeatedly to merge sorted lists [OK]

Common Mistakes:

• Appending one list after the other without sorting
• Reversing order of merged list
• Skipping nodes during merge

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

public ListNode merge(ListNode l1, ListNode l2) {
    if (l1 == null) return l2;
    if (l2 == null) return l1;
    if (l1.val < l2.val) {
        l1.next = merge(l1.next, l2);
        return l1;
    } else {
        l2.next = merge(l1, l2.next);
        return l2;
    }
}

medium

A. Missing base case for both lists being null

B. No error; code correctly merges two sorted linked lists

C. Incorrect comparison operator; should be <= instead of <

D. Function does not handle empty lists properly

Solution

1. Step 1: Check base cases

   Code returns the other list if one is null, correctly handling empty lists.

2. Step 2: Check recursive merge logic

   Compares values, recursively merges next nodes, and returns the smaller node, which is correct.

3. Final Answer:

   No error; code correctly merges two sorted linked lists -> Option B

4. Quick Check:

   Recursive merge with base cases is correct [OK]

Hint: Check base cases and recursive calls carefully [OK]

Common Mistakes:

• Forgetting to return merged list
• Incorrect base case handling
• Wrong comparison operator causing order issues

5. You have two sorted linked lists:

List 1: 1 -> 4 -> 5 -> null
List 2: 1 -> 3 -> 4 -> null

After merging, what is the correct merged list?
Also, explain why the merged list contains duplicate values.

hard

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

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

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

D. 1 -> 1 -> 4 -> 5 -> null

Solution

1. Step 1: Merge nodes in ascending order

   Compare heads: both 1, add both 1s. Then 4 vs 3, add 3. Then 4 vs 4, add both 4s. Finally add 5.

2. Step 2: Understand duplicates in merged list

   Duplicates appear because both lists contain the same values independently; merge keeps all elements without removing duplicates.

3. Final Answer:

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

4. Quick Check:

   Merge keeps all elements including duplicates [OK]

Hint: Merge keeps duplicates if present in input lists [OK]

Common Mistakes:

• Removing duplicates during merge
• Appending lists without sorting
• Skipping nodes with same values