Merge Two Sorted Linked Lists in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

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

To create a new sorted linked list that contains all elements from both lists in ascending order.

beginner

In merging two sorted linked lists, why do we compare the current nodes of both lists?

To decide which node has the smaller value and should be added next to the merged list, maintaining sorted order.

intermediate

What happens when one linked list is fully traversed before the other during merging?

The remaining nodes of the other list are appended directly to the merged list because they are already sorted.

intermediate

Why is a dummy node often used in the merge two sorted linked lists algorithm?

A dummy node simplifies the code by providing a starting point for the merged list, avoiding special cases for the head node.

beginner

What is the time complexity of merging two sorted linked lists with lengths m and n?

O(m + n) because each node from both lists is visited exactly once.

What is the first step in merging two sorted linked lists?

A Compare the first nodes of both lists

B Create a new empty list

C Reverse both lists

D Delete duplicates

We start by comparing the first nodes to decide which node to add first to the merged list.

What do you do when the current node of list1 is smaller than the current node of list2?

A Add list2's node to merged list

B Add list1's node to merged list

C Skip both nodes

D Stop merging

Add the smaller node (list1's node) to keep the merged list sorted.

If list1 is empty but list2 still has nodes, what should you do?

A Reverse list2

B Discard list2's nodes

C Start over

D Append list2's remaining nodes to merged list

Since list2 is sorted, append its remaining nodes directly.

Why is a dummy node helpful in merging two sorted linked lists?

A It stores the largest value

B It deletes duplicates

C It helps avoid special cases for the head node

D It reverses the list

A dummy node provides a fixed starting point, simplifying pointer handling.

What is the overall time complexity of merging two sorted linked lists?

A O(m + n)

B O(m * n)

C O(m^2)

D O(log(m + n))

Each node is visited once, so the time is proportional to the sum of lengths.

Explain step-by-step how to merge two sorted linked lists into one sorted linked list.

Think about walking through both lists together and picking the smaller values.

Describe why using a dummy node simplifies the merging process of two sorted linked lists.

Imagine building a list without worrying about the first node.

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