Reverse a Singly Linked List Iterative in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the previous pointer to nil.

DSA Go

var prev *Node = [1]

Drag options to blanks, or click blank then click option'

Anil

Bhead

Cnew(Node)

D&Node{}

Attempts:

3 left

💡 Hint

Common Mistakes

Setting prev to head instead of nil.

Using new(Node) which creates a new empty node.

We start with prev as nil because initially there is no previous node.

2fill in blank

medium

Complete the code to store the next node before changing links.

DSA Go

next := current[1]

Drag options to blanks, or click blank then click option'

A.nextNode

B.prev

C.next

D.link

Attempts:

3 left

💡 Hint

Common Mistakes

Using .prev which does not exist in singly linked lists.

Using incorrect field names like .nextNode or .link.

We save current.next in next to not lose the rest of the list when reversing the link.

3fill in blank

hard

Fix the error in reversing the link by assigning the correct pointer.

DSA Go

current[1] = prev

Drag options to blanks, or click blank then click option'

A.next

B.pointer

C.link

D.prev

Attempts:

3 left

💡 Hint

Common Mistakes

Trying to assign to a non-existent field like .prev.

Using wrong field names like .link or .pointer.

We reverse the link by setting current.next = prev, pointing current node back to previous.

4fill in blank

hard

Fill both blanks to update pointers for the next iteration.

DSA Go

prev, current = [1], [2]

Drag options to blanks, or click blank then click option'

Acurrent

Bnext

Cprev

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping the order incorrectly.

Using head instead of next for current.

After reversing the link, move prev to current and current to next to continue.

5fill in blank

hard

Fill all three blanks to complete the iterative reverse function.

DSA Go

func reverseList(head *Node) *Node {
    var prev *Node = [1]
    current := head
    for current != nil {
        next := current[2]
        current[3] = prev
        prev, current = current, next
    }
    return prev
}

Drag options to blanks, or click blank then click option'

Anil

B.next

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Not initializing prev to nil.

Using wrong field names for next pointer.

Forgetting to update pointers after reversing.

The function initializes prev to nil, saves current.next in next, reverses the link by current.next = prev, then moves pointers forward.

Practice

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

easy

A. To sort the linked list in ascending order

B. To create a new linked list with reversed elements

C. To delete all nodes and recreate them in reverse order

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

Solution

1. Step 1: Understand the linked list structure

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

2. Step 2: Purpose of iterative reversal

   The iterative method changes the next pointer of each node to point to the previous node, reversing the list in place.

3. Final Answer:

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

4. Quick Check:

   Reversing means changing links direction [OK]

Hint: Reversing means flipping links, not creating new nodes [OK]

Common Mistakes:

• Thinking it creates a new list instead of reversing in place
• Confusing reversal with sorting
• Assuming nodes are deleted and recreated

2. Which of the following Go code snippets correctly initializes pointers for iterative reversal of a singly linked list?

easy

A. var prev = head; current := nil

B. var prev *Node = nil; current := head

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

D. var prev *Node = head; current := head

Solution

1. Step 1: Initialization for reversal

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

2. Step 2: Check each option

   var prev *Node = nil; current := head correctly sets prev to nil and current to head. Others incorrectly assign prev or current.

3. Final Answer:

   var prev *Node = nil; current := head -> Option B

4. Quick Check:

   Prev starts nil, current at head [OK]

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

Common Mistakes:

• Setting prev to head initially
• Setting current to nil at start
• Mixing pointer assignments

3. What will be the output linked list after running this Go code to reverse a singly linked list?

type Node struct {
    Val  int
    Next *Node
}

func reverseList(head *Node) *Node {
    var prev *Node = nil
    current := head
    for current != nil {
        nextTemp := current.Next
        current.Next = prev
        prev = current
        current = nextTemp
    }
    return prev
}

// Input list: 1 -> 2 -> 3 -> nil
// Call: reverseList(head)

medium

A. 3 -> 2 -> 1 -> nil

B. 1 -> 2 -> 3 -> nil

C. nil -> 3 -> 2 -> 1

D. 3 -> 1 -> 2 -> nil

Solution

1. Step 1: Trace the reversal loop

   Initially prev=nil, current=1. Each iteration reverses current.Next to prev and moves pointers forward.

2. Step 2: Final reversed list

   After loop, prev points to node with Val=3, linked as 3->2->1->nil.

3. Final Answer:

   3 -> 2 -> 1 -> nil -> Option A

4. Quick Check:

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

Hint: Follow pointer changes step-by-step [OK]

Common Mistakes:

• Confusing original and reversed order
• Misunderstanding pointer updates
• Assuming list remains unchanged

4. Identify the error in this Go code snippet for reversing a singly linked list iteratively:

func reverseList(head *Node) *Node {
    var prev *Node = head
    current := head
    for current != nil {
        nextTemp := current.Next
        current.Next = prev
        prev = current
        current = nextTemp
    }
    return prev
}

medium

A. prev should be initialized to nil, not head

B. current should be initialized to nil

C. nextTemp should be initialized outside the loop

D. The function should return current, not prev

Solution

1. Step 1: Check initialization of prev

   Prev must start as nil because the new tail's next is nil. Setting prev=head causes a cycle.

2. Step 2: Effect of wrong prev initialization

   Setting prev=head makes current.Next point to itself on first iteration, causing infinite loop or wrong reversal.

3. Final Answer:

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

4. Quick Check:

   Prev starts nil to avoid cycles [OK]

Hint: Prev must start nil to break old links [OK]

Common Mistakes:

• Initializing prev to head causing cycles
• Returning current instead of prev
• Misplacing nextTemp initialization

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

hard

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

B. nil

C. 10 -> 20 -> 30 -> 40 -> nil

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

Solution

1. Step 1: First reversal

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

2. Step 2: Second reversal

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

3. Final Answer:

   10 -> 20 -> 30 -> 40 -> nil -> Option C

4. Quick Check:

   Double reversal returns original list [OK]

Hint: Reversing twice returns original list [OK]

Common Mistakes:

• Assuming second reversal changes order differently
• Confusing node values after reversal
• Thinking list becomes empty