Merge Two Sorted Linked Lists in DSA Java

Mental Model

We combine two sorted lists into one sorted list by picking the smallest current item from either list step by step.

Analogy: Imagine merging two lines of people waiting for a ride, each line sorted by height. You pick the shortest person from the front of either line to form a new line sorted by height.

List1: 1 -> 3 -> 5 -> null
List2: 2 -> 4 -> 6 -> null
Merged: null

Dry Run Walkthrough

Input: list1: 1->3->5, list2: 2->4->6

Goal: Create one sorted list containing all elements from both lists

Step 1: Compare heads: 1 (list1) and 2 (list2), pick 1

Merged: 1 -> null
List1: 3 -> 5 -> null
List2: 2 -> 4 -> 6 -> null

Why: 1 is smaller, so it should come first in merged list

Step 2: Compare heads: 3 (list1) and 2 (list2), pick 2

Merged: 1 -> 2 -> null
List1: 3 -> 5 -> null
List2: 4 -> 6 -> null

Why: 2 is smaller, so it comes next

Step 3: Compare heads: 3 (list1) and 4 (list2), pick 3

Merged: 1 -> 2 -> 3 -> null
List1: 5 -> null
List2: 4 -> 6 -> null

Why: 3 is smaller, so it comes next

Step 4: Compare heads: 5 (list1) and 4 (list2), pick 4

Merged: 1 -> 2 -> 3 -> 4 -> null
List1: 5 -> null
List2: 6 -> null

Why: 4 is smaller, so it comes next

Step 5: Compare heads: 5 (list1) and 6 (list2), pick 5

Merged: 1 -> 2 -> 3 -> 4 -> 5 -> null
List1: null
List2: 6 -> null

Why: 5 is smaller, so it comes next

Step 6: List1 empty, append remaining list2 nodes

Merged: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null
List1: null
List2: null

Why: No more nodes in list1, so add all remaining nodes from list2

Result:

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

Annotated Code

DSA Java

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

public class MergeSortedLists {
    public static ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0); // dummy node to start merged list
        ListNode current = dummy; // pointer to build merged list

        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                current.next = l1; // attach l1 node
                l1 = l1.next; // move l1 forward
            } else {
                current.next = l2; // attach l2 node
                l2 = l2.next; // move l2 forward
            }
            current = current.next; // move current forward
        }

        if (l1 != null) {
            current.next = l1; // attach remaining l1 nodes
        } else {
            current.next = l2; // attach remaining l2 nodes
        }

        return dummy.next; // head of merged list
    }

    public static void printList(ListNode head) {
        ListNode curr = head;
        while (curr != null) {
            System.out.print(curr.val);
            if (curr.next != null) System.out.print(" -> ");
            curr = curr.next;
        }
        System.out.println(" -> null");
    }

    public static void main(String[] args) {
        ListNode l1 = new ListNode(1);
        l1.next = new ListNode(3);
        l1.next.next = new ListNode(5);

        ListNode l2 = new ListNode(2);
        l2.next = new ListNode(4);
        l2.next.next = new ListNode(6);

        ListNode merged = mergeTwoLists(l1, l2);
        printList(merged);
    }
}

while (l1 != null && l2 != null) {

continue merging while both lists have nodes

if (l1.val < l2.val) { current.next = l1; l1 = l1.next; } else { current.next = l2; l2 = l2.next; }

pick smaller node and advance that list

current = current.next;

advance current pointer to keep building merged list

if (l1 != null) { current.next = l1; } else { current.next = l2; }

attach remaining nodes from non-empty list

return dummy.next;

return head of merged list skipping dummy

Output

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

Complexity Analysis

Time: O(n + m) because we traverse each node in both lists once

Space: O(1) because we rearrange pointers without extra storage

vs Alternative: Compared to merging into a new list with new nodes, this approach reuses existing nodes saving space and time

Edge Cases

One or both lists empty

Returns the non-empty list or null if both empty

DSA Java

while (l1 != null && l2 != null) {

Lists with duplicate values

Duplicates are merged maintaining sorted order

DSA Java

if (l1.val < l2.val) { ... } else { ... }

Lists of different lengths

Remaining nodes of longer list appended after shorter list ends

DSA Java

if (l1 != null) { current.next = l1; } else { current.next = l2; }

When to Use This Pattern

When you see two sorted lists and need a combined sorted list, use the merge pattern by comparing heads stepwise to build the result.

Common Mistakes

Mistake: Not advancing the current pointer after attaching a node
Fix: Add 'current = current.next;' after attaching a node to keep building the merged list

Mistake: Forgetting to attach remaining nodes after one list ends
Fix: After the loop, attach the non-empty list to current.next

Summary

It merges two sorted linked lists into one sorted linked list by choosing the smaller head node repeatedly.

Use it when you have two sorted lists and want one sorted combined list efficiently.

The key insight is to pick the smaller current node from either list and advance pointers until all nodes are merged.

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