Reverse a Singly Linked List Iterative in DSA Java - Time & Space Complexity

Time Complexity: Reverse a Singly Linked List Iterative

O(n)

Understanding Time Complexity

We want to understand how the time needed to reverse a singly linked list changes as the list grows.

How does the number of steps grow when the list gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public ListNode reverseList(ListNode head) {
    ListNode prev = null;
    ListNode current = head;
    while (current != null) {
        ListNode nextTemp = current.next;
        current.next = prev;
        prev = current;
        current = nextTemp;
    }
    return prev;
}
    

This code reverses a singly linked list by changing the direction of the links one by one.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that visits each node once.
• How many times: Exactly once for each node in the list (n times).

How Execution Grows With Input

As the list gets longer, the number of steps grows directly with the number of nodes.

Input Size (n)	Approx. Operations
10	About 10 steps
100	About 100 steps
1000	About 1000 steps

Pattern observation: The steps increase in a straight line as the list size grows.

Final Time Complexity

Time Complexity: O(n)

This means the time to reverse the list grows directly with the number of nodes.

Common Mistake

[X] Wrong: "Reversing a linked list takes more than linear time because of pointer changes."

[OK] Correct: Each node is visited once and its pointer changed once, so the work grows only linearly with the list size.

Interview Connect

Understanding this time complexity helps you explain why reversing a list is efficient and expected to be done in one pass.

Self-Check

"What if we used recursion instead of iteration? How would the time complexity change?"

Practice

1. What is the main purpose of reversing a singly linked list iteratively?

easy

A. To sort the list in ascending order

B. To delete all nodes from the list

C. To change the direction of all links so the last node becomes the first

D. To add a new node at the end of the list

Solution

1. Step 1: Understand the linked list structure

   A singly linked list has nodes connected in one direction from head to tail.

2. Step 2: Purpose of reversing

   Reversing changes the links so the last node points to the previous one, making it the new head.

3. Final Answer:

   To change the direction of all links so the last node becomes the first -> Option C

4. Quick Check:

   Reversing = changing link directions [OK]

Hint: Reversing flips the list direction, last becomes first [OK]

Common Mistakes:

• Confusing reversing with sorting
• Thinking reversing deletes nodes
• Assuming reversing adds nodes

2. Which of the following is the correct way to initialize the previous pointer before starting the iterative reversal?

easy

A. Node prev = head;

B. Node prev = head.next;

C. Node prev = new Node();

D. Node prev = null;

Solution

1. Step 1: Understand the role of previous pointer

   Previous pointer starts as null because the new tail's next should be null after reversal.

2. Step 2: Initialization

   Setting prev to null ensures the first node processed points to null, ending the reversed list properly.

3. Final Answer:

   Node prev = null; -> Option D

4. Quick Check:

   Prev starts null to mark new tail [OK]

Hint: Prev pointer starts null to mark list end [OK]

Common Mistakes:

• Initializing prev to head causes cycle
• Creating new node wastes memory
• Starting prev at head.next skips first node

3. What will be the output linked list after reversing the list 1 -> 2 -> 3 -> null iteratively?

medium

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

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

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

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

Solution

1. Step 1: Trace the reversal process

   Initially: 1 -> 2 -> 3 -> null. After reversal, links point backward: 3 -> 2 -> 1 -> null.

2. Step 2: Confirm final head and tail

   Node 3 becomes new head, node 1 becomes tail pointing to null.

3. Final Answer:

   3 -> 2 -> 1 -> null -> Option B

4. Quick Check:

   Reversed list = 3 -> 2 -> 1 [OK]

Hint: Last node becomes first after reversal [OK]

Common Mistakes:

• Assuming list stays same after reversal
• Confusing null placement
• Mixing node order

4. Identify the error in this iterative reversal code snippet:

Node prev = head;
Node current = head.next;
while (current != null) {
  Node nextNode = current.next;
  current.next = prev;
  prev = current;
  current = nextNode;
}
head.next = null;
head = prev;

medium

A. prev should be initialized to null, not head

B. current should start at head, not head.next

C. head.next should not be set to null

D. Loop condition should be current.next != null

Solution

1. Step 1: Check initialization of prev pointer

   Prev must start as null to mark new tail's next as null after reversal.

2. Step 2: Effect of wrong initialization

   Starting prev at head causes a cycle, breaking reversal logic.

3. Final Answer:

   prev should be initialized to null, not head -> Option A

4. Quick Check:

   Prev = null to avoid cycles [OK]

Hint: Prev must start null to avoid cycles [OK]

Common Mistakes:

• Starting prev at head causes infinite loop
• Wrong loop condition misses last node
• Forgetting to set old head next to null

5. Given a singly linked list with nodes containing values 10 -> 20 -> 30 -> 40 -> null, which iterative code snippet correctly reverses it and prints the reversed list?

hard

A. Node prev = null;
Node current = head;
while (current != null) {
Node next = current.next;
current.next = prev;
prev = current;
current = next;
}
head = prev;
// Print list starting from head

B. Node prev = head;
Node current = head.next;
while (current != null) {
Node next = current.next;
current.next = prev;
prev = current;
current = next;
}
head = prev;
// Print list starting from head

C. Node current = head;
while (current != null) {
Node next = current.next;
current.next = null;
current = next;
}
// Print list starting from head

D. Node prev = null;
Node current = head;
while (current.next != null) {
Node next = current.next;
current.next = prev;
prev = current;
current = next;
}
head = prev;
// Print list starting from head

Solution

1. Step 1: Verify initialization of pointers

   Prev must start as null, current at head to reverse all nodes correctly.

2. Step 2: Check loop condition and pointer updates

   Loop runs while current != null to process all nodes; inside loop, links reversed properly.

3. Step 3: Confirm final head assignment and printing

   After loop, head is set to prev (new head). Printing from head shows reversed list.

4. Final Answer:

   Option A code correctly reverses and prints the list -> Option A

5. Quick Check:

   Prev=null, loop full list, head=prev [OK]

Hint: Prev null, loop full list, head = prev after reversal [OK]

Common Mistakes:

• Starting prev at head causes cycle
• Loop condition misses last node
• Setting current.next = null inside loop breaks links