Reverse a Singly Linked List Iterative in DSA C# - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind reversing a singly linked list iteratively?

We change the direction of each node's pointer one by one, so the list points backward instead of forward.

beginner

In the iterative reversal of a singly linked list, which pointers do we typically use?

We use three pointers: previous (initially null), current (starting at head), and next (to store current's next node).

intermediate

Why do we need to store the next node before changing the current node's next pointer?

Because once we reverse the pointer, we lose the original next node. Storing it first lets us continue traversing the list.

beginner

What is the time complexity of reversing a singly linked list iteratively?

It is O(n), where n is the number of nodes, because we visit each node exactly once.

beginner

Show the final state of the list after reversing: 1 -> 2 -> 3 -> null

After reversal, the list becomes: 3 -> 2 -> 1 -> null

Which pointer should initially be set to null when reversing a singly linked list iteratively?

A previous

B current

C next

D head

The previous pointer starts as null because the new tail node will point to null.

What happens to the current node's next pointer during each iteration?

A It is set to null

B It is set to next node

C It remains unchanged

D It is set to previous node

We reverse the pointer by setting current's next to previous.

Why do we need a 'next' pointer in the iterative reversal?

A To mark the end of the list

B To keep track of the next node before changing pointers

C To store the previous node

D To count the nodes

We save the next node to continue traversal after reversing the pointer.

What is the new head of the list after reversal?

A The original tail node

B The original head node

C Null

D The middle node

After reversal, the original tail becomes the new head.

What is the space complexity of iterative reversal of a singly linked list?

A O(log n)

B O(n)

C O(1)

D O(n^2)

We only use a few pointers, so space complexity is constant.

Explain step-by-step how to reverse a singly linked list using iteration.

Think about how pointers move and change direction.

Describe why storing the next node before reversing pointers is important in iterative reversal.

Imagine walking backward but needing to know where to go next.

Practice

1. What is the main purpose of the iterative method to reverse a singly linked list?

easy

A. To sort the nodes in ascending order

B. To delete all nodes from the list

C. To change the direction of the links between nodes one by one

D. To add new nodes 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 through next pointers.

2. Step 2: Purpose of iterative reversal

   The iterative method changes each node's next pointer to point backward, reversing the list.

3. Final Answer:

   To change the direction of the links between nodes one by one -> Option C

4. Quick Check:

   Reversing = changing links direction [OK]

Hint: Reversing means flipping the links one by one [OK]

Common Mistakes:

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

2. Which of the following is the correct way to initialize pointers for iterative reversal of a singly linked list in C#?

easy

A. Node prev = head; Node current = null;

B. Node prev = null; Node current = null;

C. Node prev = head; Node current = head.next;

D. Node prev = null; Node current = head;

Solution

1. Step 1: Initialization for reversal

   We start with prev as null because the new tail will point to null, and current at head to traverse.

2. Step 2: Check each option

   Only Node prev = null; Node current = head; correctly sets prev to null and current to head for iteration.

3. Final Answer:

   Node prev = null; Node current = head; -> Option D

4. Quick Check:

   Start prev=null, current=head [OK]

Hint: Prev starts null, current starts at head [OK]

Common Mistakes:

• Setting prev to head initially
• Setting current to null
• Both pointers null at start

3. What will be the output linked list after running this code snippet on list 1 -> 2 -> 3 -> null?

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

medium

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

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

C. null

D. 1 -> null

Solution

1. Step 1: Trace the reversal loop

   Each iteration reverses the link: 1 points to null, 2 points to 1, 3 points to 2.

2. Step 2: Final linked list after loop

   After loop, prev points to 3, which is new head: 3 -> 2 -> 1 -> null.

3. Final Answer:

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

4. Quick Check:

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

Hint: Reversal flips list order completely [OK]

Common Mistakes:

• Assuming list stays same
• Forgetting to update head
• Confusing next and prev pointers

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

Node prev = head;
Node current = head.next;
while (current != null) {
  Node nextTemp = current.next;
  current.next = prev;
  prev = current;
  current = nextTemp;
}
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. No error, code is correct

Solution

1. Step 1: Check pointer initialization

   prev must start as null to mark new tail's next as null; starting at head causes cycle.

2. Step 2: Effect of wrong initialization

   Starting prev at head causes the first node to point to itself, creating a loop.

3. Final Answer:

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

4. Quick Check:

   Prev must be null at start [OK]

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

Common Mistakes:

• Starting prev at head causing loops
• Not setting head.next to null after reversal
• Starting current at wrong node

5. Given a singly linked list with nodes containing values 10 -> 20 -> 30 -> 40 -> null, write the final linked list after applying iterative reversal twice in a row.

hard

A. 10 -> 20 -> 30 -> 40 -> null

B. 40 -> 30 -> 20 -> 10 -> null

C. null

D. 40 -> 20 -> 30 -> 10 -> null

Solution

1. Step 1: First reversal

   Reversing 10->20->30->40 gives 40->30->20->10->null.

2. Step 2: Second reversal

   Reversing again restores original order: 10->20->30->40->null.

3. Final Answer:

   10 -> 20 -> 30 -> 40 -> null -> Option A

4. Quick Check:

   Double reversal restores original list [OK]

Hint: Reversing twice returns original list [OK]

Common Mistakes:

• Assuming second reversal changes order again
• Confusing node values after reversal
• Forgetting reversal is reversible