Reverse a Singly Linked List Iterative in DSA Go - 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 between nodes one by one, using three pointers: previous, current, and next. This way, we reverse the list without creating new nodes.

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 before the current node once the link is reversed. It starts as nil and moves forward as we reverse links.

beginner

Why do we need a 'next' pointer during the iterative reversal of a singly linked list?

Because once we reverse the current node's link, we lose the original next node. The 'next' pointer saves the original next node before reversing the link, so we can continue traversing.

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

What will be the new head of the list after reversing a singly linked list iteratively?

The new head will be the last node of the original list, which is tracked by the 'previous' pointer after the iteration finishes.

Which pointer is used to keep track of the next node before reversing the link?

A next

B previous

C current

D head

The 'next' pointer stores the original next node before reversing the current node's link.

What is the initial value of the 'previous' pointer when starting the iterative reversal?

A head

B tail

C nil (or null)

D current

The 'previous' pointer starts as nil because there is no node before the head initially.

After reversing the list, which pointer points to the new head?

A previous

B next

C current

D tail

The 'previous' pointer ends up pointing to the new head after all links are reversed.

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

A O(1)

B O(n)

C O(n^2)

D O(log n)

We visit each node once, so the time complexity is O(n).

Which of these is NOT needed to reverse a singly linked list iteratively?

A previous pointer

B current pointer

C next pointer

D extra array to store nodes

No extra array is needed; reversal is done by changing links using pointers.

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

Think about how pointers move and links change in each step.

Why is it important to save the next node before changing the current node's link during reversal?

Consider what happens if you reverse a link without saving the next node.

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