Reverse a Singly Linked List Iterative in DSA C# - 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 gets bigger.

Specifically, how many steps does it take to reverse all the nodes?

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 the list takes more time because of multiple pointer changes inside the loop."

[OK] Correct: Each node is visited only once, and pointer changes are constant time operations, so the total time still grows linearly.

Interview Connect

Understanding this linear time reversal is a key skill that shows you can work with linked lists efficiently and confidently.

Self-Check

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

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