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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the previous pointer to null.

DSA Javascript

let prev = [1];

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

A0

Bnull

Cundefined

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Setting prev to head instead of null

Using undefined instead of null

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

2fill in blank

medium

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

DSA Javascript

let next = [1].next;

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

Acurrent

Bprev

Chead

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using prev.next instead of current.next

Using next.next which is undefined

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

3fill in blank

hard

Fix the error in reversing the link by assigning the previous node to current's next.

DSA Javascript

[1].next = prev;

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

Anext

Bprev

Ccurrent

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning next.next instead of current.next

Assigning prev.next which is wrong

We reverse the link by pointing current.next to prev.

4fill in blank

hard

Fill both blanks to update pointers for the next iteration.

DSA Javascript

prev = [1];
current = [2];

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

Acurrent

Bprev

Cnext

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping the order of assignments

Using prev for current instead of next

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 Javascript

function reverseList(head) {
  let prev = [1];
  let current = head;
  while (current !== null) {
    let next = current.[2];
    current.[3] = prev;
    prev = current;
    current = next;
  }
  return prev;
}

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

Anull

Bnext

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable names

Forgetting to return prev

Mixing up next and prev

Initialize prev to null. Save current.next in next. Reverse link by setting current.next = prev.

Practice

1. What does reversing a singly linked list iteratively mean?
head -> 1 -> 2 -> 3 -> null becomes?

easy

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

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

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

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

Solution

1. Step 1: Understand the linked list order

   The original list goes from 1 to 3 in order: 1 -> 2 -> 3 -> null.

2. Step 2: Reverse the pointers iteratively

   Reversing means the last node becomes first, so the list becomes 3 -> 2 -> 1 -> null.

3. Final Answer:

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

4. Quick Check:

   Reverse order = head -> 3 -> 2 -> 1 -> null [OK]

Hint: Reversing flips the list order completely [OK]

Common Mistakes:

• Thinking the list stays the same after reverse
• Confusing node values with pointers
• Assuming null moves to front

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

easy

A. let prev = null, current = null;

B. let prev = head, current = null;

C. let prev = head, current = head.next;

D. let prev = null, current = head;

Solution

1. Step 1: Initialize previous pointer to null

   Before reversal, previous node is null because head has no previous.

2. Step 2: Initialize current pointer to head

   Current starts at head to traverse the list.

3. Final Answer:

   let prev = null, 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 at start
• Swapping initialization order

3. What is the output of the following code?

class Node {
  constructor(val) {
    this.val = val;
    this.next = null;
  }
}

function reverseList(head) {
  let prev = null;
  let current = head;
  while (current) {
    let nextTemp = current.next;
    current.next = prev;
    prev = current;
    current = nextTemp;
  }
  return prev;
}

// Create list: 1 -> 2 -> 3 -> null
let head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);

let reversedHead = reverseList(head);

let result = [];
let node = reversedHead;
while (node) {
  result.push(node.val);
  node = node.next;
}
console.log(result);

medium

A. []

B. [3, 2, 1]

C. [null]

D. [1, 2, 3]

Solution

1. Step 1: Trace reversal loop

   Each iteration reverses the next pointer: 1->2->3 becomes 3->2->1.

2. Step 2: Collect values from reversed list

   After reversal, traversing from prev (new head) gives values 3, 2, 1.

3. Final Answer:

   [3, 2, 1] -> Option B

4. Quick Check:

   Reversed list values = [3, 2, 1] [OK]

Hint: Reversed list values appear in reverse order [OK]

Common Mistakes:

• Returning original head instead of prev
• Not updating current pointer correctly
• Collecting values before reversal

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

function reverseList(head) {
  let prev = head;
  let current = head.next;
  while (current !== null) {
    let nextTemp = current.next;
    current.next = prev;
    prev = current;
    current = nextTemp;
  }
  head.next = null;
  return prev;
}

medium

A. head.next should not be set to null

B. current should start as head, not head.next

C. prev should start as null, not head

D. nextTemp should be assigned after changing current.next

Solution

1. Step 1: Check initial pointer setup

   prev must start as null because head has no previous node before reversal.

2. Step 2: Understand impact of wrong prev initialization

   Starting prev as head causes a cycle or incorrect links during reversal.

3. Final Answer:

   prev should start as null, not head -> Option C

4. Quick Check:

   prev=null start fixes reversal [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
• Changing nextTemp after modifying current.next

5. Given a singly linked list with nodes containing values 1 -> 2 -> 3 -> 4 -> 5 -> null, write an iterative function to reverse only the first 3 nodes. What will be the list after reversal?

hard

A. 3 -> 2 -> 1 -> 4 -> 5 -> null

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

C. 5 -> 4 -> 3 -> 2 -> 1 -> null

D. 1 -> 2 -> 3 -> 4 -> 5 -> null

Solution

1. Step 1: Reverse first 3 nodes iteratively

   Reverse nodes 1, 2, 3 to get 3 -> 2 -> 1.

2. Step 2: Connect reversed part to remaining list

   Node 1 (now last in reversed part) points to node 4, preserving rest of list.

3. Final Answer:

   3 -> 2 -> 1 -> 4 -> 5 -> null -> Option A

4. Quick Check:

   Partial reversal keeps rest intact [OK]

Hint: Reverse only first k nodes, link rest after [OK]

Common Mistakes:

• Reversing entire list instead of first 3 nodes
• Not reconnecting reversed part to rest
• Breaking list by losing tail connections