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

Recall & Review

beginner

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

We change the direction of the links one by one, so each node points to its previous node instead of the next. We use three pointers: previous, current, and next to keep track while moving through the list.

beginner

In the iterative reversal of a singly linked list, what role does the 'previous' pointer play?

The 'previous' pointer keeps track of the node that should come after the current node once the link is reversed. It starts as None and moves forward as we reverse links.

intermediate

Why do we need to save the 'next' node before changing the current node's next pointer during reversal?

Because once we change the current node's next pointer to point backward, we lose the original forward link. Saving 'next' lets us continue traversing the list without losing nodes.

beginner

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

The time complexity is O(n), where n is the number of nodes, because we visit each node exactly once to reverse the links.

beginner

Show the final state of the linked list after reversing 1 -> 2 -> 3 -> null iteratively.

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

Which pointer should initially be set to None when reversing a singly linked list iteratively?

A previous

B current

C next

D head

The 'previous' pointer starts as None because the new tail of the reversed list will point to None.

What must you do before changing the current node's next pointer during reversal?

A Delete the current node

B Set previous to None

C Move current to previous

D Save the next node

Saving the next node is necessary to continue traversal after changing the current node's next pointer.

What is the new head of the list after iterative reversal?

A The original tail node

B The original head node

C The middle node

D None

After reversal, the original tail becomes the new head.

How many times do we visit each node during iterative reversal?

A Twice

B Once

C Three times

D Zero times

Each node is visited exactly once to reverse the links.

What happens if you forget to update the 'previous' pointer during iteration?

A The list will become empty

B The list will reverse correctly

C The list will not reverse correctly

D The list will double in size

Not updating 'previous' breaks the chain of reversed links, so reversal fails.

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

Think about how pointers move and change direction.

Describe why saving the next node before reversing the link is crucial.

Imagine losing your map while walking backwards.

Practice

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

easy

A. To remove duplicate nodes from the list

B. To sort the list in ascending order

C. To change the direction of all links so the list order is reversed

D. To find the middle node of the list

Solution

1. Step 1: Understand the linked list structure

   A singly linked list has nodes pointing only to the next node, so reversing means changing these pointers to point backwards.

2. Step 2: Purpose of iterative reversal

   The iterative method changes each node's next pointer to the previous node, effectively reversing the list order.

3. Final Answer:

   To change the direction of all links so the list order is reversed -> Option C

4. Quick Check:

   Reversing means flipping links direction [OK]

Hint: Reversing flips all next pointers to previous nodes [OK]

Common Mistakes:

• Confusing reversal with sorting
• Thinking reversal deletes nodes
• Assuming reversal finds middle node

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

easy

A. prev = head; current = head.next

B. prev = null; current = head

C. prev = head; current = null

D. prev = null; current = null

Solution

1. Step 1: Understand pointer roles

   Prev starts as null because the new tail will point to null; current starts at head to traverse the list.

2. Step 2: Correct initialization

   Setting prev = null and current = head allows iteration from the start with no previous node initially.

3. Final Answer:

   prev = null; current = head -> Option B

4. Quick Check:

   Prev null, current at head to start [OK]

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

Common Mistakes:

• Setting prev to head initially
• Starting current as null
• Both pointers as null

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

medium

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

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

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

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

Solution

1. Step 1: Trace reversal process

   Initially: prev=null, current=1. After first iteration, 1 points to null. Then 2 points to 1, then 3 points to 2.

2. Step 2: Final reversed list

   The list becomes 3 -> 2 -> 1 -> null after all nodes reverse their next pointers.

3. Final Answer:

   3 -> 2 -> 1 -> null -> Option A

4. Quick Check:

   Reversed list order is 3, 2, 1 [OK]

Hint: Last node becomes new head after reversal [OK]

Common Mistakes:

• Not updating pointers correctly
• Confusing original and reversed order
• Including null as a node

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

def reverse(head):
    prev = head
    current = null
    while current:
        next_node = current.next
        current.next = prev
        prev = current
        current = next_node
    return prev

medium

A. prev should start as null, current as head

B. The loop condition should be 'while prev' instead of 'while current'

C. current.next assignment is reversed, should be 'prev.next = current'

D. Function should return current instead of prev

Solution

1. Step 1: Check pointer initialization

   Prev should be null initially because the new tail points to null; current should start at head to traverse.

2. Step 2: Loop condition and pointer updates

   Loop runs while current is not null; current.next is set to prev correctly; returning prev gives new head.

3. Final Answer:

   prev should start as null, current as head -> Option A

4. Quick Check:

   Prev null, current head to start loop [OK]

Hint: Prev null, current head to avoid skipping nodes [OK]

Common Mistakes:

• Starting prev at head causes cycle
• Loop condition incorrect
• Returning wrong pointer

5. Given a singly linked list with nodes containing values [10, 20, 30, 40, 50], which iterative reversal step correctly updates the pointers after processing the first two nodes?

hard

A. After first node: 10 -> 20; After second node: 20 -> null

B. After first node: 10 -> 20 -> null; After second node: 20 -> 10 -> null

C. After first node: null -> 10; After second node: 10 -> 20

D. After first node: 10 -> null; After second node: 20 -> 10 -> null

Solution

1. Step 1: Reverse first node pointer

   Initially, prev=null, current=10. After first iteration, 10.next = null, so list is 10 -> null.

2. Step 2: Reverse second node pointer

   Next, current=20. Set 20.next = prev (which is 10), so list is 20 -> 10 -> null.

3. Final Answer:

   After first node: 10 -> null; After second node: 20 -> 10 -> null -> Option D

4. Quick Check:

   Each node points back to previous [OK]

Hint: Each step points current node to previous node [OK]

Common Mistakes:

• Not setting first node's next to null
• Linking nodes forward instead of backward
• Confusing order of pointer updates