Create a Circular Singly Linked List in DSA Rust - Execution Trace

Concept Flow - Create a Circular Singly Linked List

Create new node

↓

Is list empty?

No→Traverse to last node

↓

Set last.next = new node

↓

Set head = new node

↓

Set new node.next = head

↓

Done

Start by creating a new node. If the list is empty, set head to new node and point new node to itself. Otherwise, find the last node, link it to new node, and link new node back to head.

Execution Sample

DSA Rust

use std::ptr;

struct Node {
    data: i32,
    next: *mut Node,
}

// Create circular list with one node
let mut head: *mut Node = ptr::null_mut();
let new_node = Box::new(Node { data: 10, next: ptr::null_mut() });
let new_node_ptr = Box::into_raw(new_node);
head = new_node_ptr;
unsafe {
    (*head).next = head;
}

This code creates a circular singly linked list with one node pointing to itself using raw pointers, which is a common approach for DSA in Rust to allow cycles.

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	Create new node with data=10	new_node_ptr -> Node(10)	new_node_ptr -> Node(10), head = null_mut	Node(10) -> null_mut
2	Check if list is empty (head.is_null())	new_node_ptr -> Node(10)	head.is_null() == true	Node(10) -> null_mut
3	Set head = new_node_ptr	head -> Node(10)	head -> new_node_ptr	head -> Node(10) -> null_mut
4	Set (*head).next = head (making circular)	head -> Node(10)	(*head).next -> head	head -> Node(10) -> head (circular)

💡 List created with one node pointing to itself, forming a circular singly linked list.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	Final
head	null_mut	null_mut	null_mut	Node(10)	Node(10)
Node(10).next	N/A	null_mut	null_mut	null_mut	Node(10)

Key Moments - 3 Insights

Why do we set the new node's next pointer to head when creating a circular list?

Setting the new node's next to head creates the circular link, so traversal from any node loops back to the start. See execution_table step 4 where (*head).next points to head.

What happens if the list is empty when adding the first node?

If the list is empty, we set head to the new node_ptr and then point new node's next to itself, making a single-node circular list. This is shown in execution_table steps 2, 3 and 4.

Why does the new node point to itself in a single-node circular list?

Because in a circular list, the last node's next points back to head. With one node, head is that node, so it points to itself. See visual state in execution_table step 4.

Visual Quiz - 3 Questions

Look at the execution_table, what does Node(10).next point to after step 4?

A A new node

B null_mut

C head (Node(10))

D Null pointer

💡 Hint

Check the 'Pointer Changes' and 'Visual State' columns in step 4.

At which step is the list confirmed to be empty?

A Step 2

B Step 3

C Step 1

D Step 4

💡 Hint

Look at the 'Operation' column where the list emptiness is checked.

If we did not set the new node's next to head, what would the visual state be after step 4?

A Node(10) -> Node(10)

B Node(10) -> null_mut

C head -> null_mut

D Empty list

💡 Hint

Refer to the 'Visual State' column in step 3.

Concept Snapshot

Create a new node.
If list empty: head = new_node_ptr, (*new_node_ptr).next = head (self).
Else: find last node, last.next = new_node_ptr, (*new_node_ptr).next = head.
This forms a circular singly linked list where last node points back to head.

Full Transcript

To create a circular singly linked list, start by making a new node. If the list is empty, set head to this new node and point its next to itself, making a single-node circle. If the list has nodes, find the last node, link it to the new node, and then link the new node back to head. This way, the list loops back to the start, forming a circle. The key step is setting the new node's next pointer to head, which closes the loop. This process ensures traversal never ends and cycles through the nodes continuously.

Practice

1. What distinguishes a circular singly linked list from a regular singly linked list?

easy

A. Circular lists have two pointers per node

B. The last node's next pointer points back to the head instead of None

C. Circular lists store data in reverse order

D. Each node stores a reference to both its predecessor and successor

Solution

1. Step 1: Regular linked list

   In a regular singly linked list, the last node's next is None - marking the end.

2. Step 2: Circular difference

   In a circular singly linked list, the last node's next points back to the head. There is no None terminator - traversal wraps around.

3. Final Answer:

   Last node points back to head. -> Option B

4. Quick Check:

   Circular = last.next → head (not None). [OK]

Hint: Circular = last node's next points to head, not None. [OK]

Common Mistakes:

• Confusing with doubly linked list (two pointers per node)
• Thinking circular means reverse storage
• Confusing circular with bidirectional

2. What is the time complexity of inserting a node at the end of a circular singly linked list when you maintain a tail pointer?

easy

A. O(n)

B. O(log n)

C. O(1)

D. O(n²)

Solution

1. Step 1: With tail pointer

   tail.next = new_node (new last). new_node.next = head (circular link). tail = new_node. All O(1) pointer updates.

2. Step 2: Without tail pointer

   Would need O(n) traversal to find the last node.

3. Final Answer:

   O(1) with tail pointer maintained. -> Option C

4. Quick Check:

   Tail pointer = O(1) insert at end. Without tail = O(n). [OK]

Hint: Tail pointer = O(1) end insertion. Always maintain tail for circular lists. [OK]

Common Mistakes:

• Saying O(n) because traversal is needed without tail pointer
• Not knowing tail pointer eliminates the traversal
• Confusing with insertion at arbitrary position which is O(n)

3. In Rust, why is creating a circular linked list with safe Box<T> impossible without additional indirection?

medium

A. Box<T> doesn't support linked lists

B. Box<T> implies unique ownership - a circular structure requires the last node to own the first, creating a cycle that Box cannot express

C. Rust doesn't allow struct self-reference

D. Box<T> is only for heap allocation, not pointer structures

Solution

1. Step 1: Box ownership

   Box<T> gives unique ownership of heap data. The last node would need to own the head node.

2. Step 2: Circular ownership cycle

   But the head is also owned by the list itself. Two owners of the same node = illegal in Rust. Box cannot express shared or circular ownership.

3. Step 3: Solution

   Use Rc for shared ownership, or raw pointers with unsafe, or index-based representation with Vec.

4. Final Answer:

   Box implies unique ownership - circular structure needs shared ownership (Rc) or unsafe raw pointers. -> Option B

5. Quick Check:

   Circular = shared ownership. Box = unique ownership. Mismatch. [OK]

Hint: Circular needs shared ownership. Box=unique. Use Rc> or raw pointers. [OK]

Common Mistakes:

• Thinking Box can handle any heap structure
• Not knowing Rust's ownership applies to data structure design
• Confusing Box limitations with general Rust limitations

4. How do you detect that you've completed a full traversal of a circular singly linked list?

medium

A. Check if current.next is None

B. Check if current.next points to the head node

C. Count nodes until reaching n

D. Check if current equals the tail node

Solution

1. Step 1: No None terminator

   Circular lists have no None - checking next is None never terminates traversal.

2. Step 2: Termination condition

   Start at head. Advance. Stop when current.next == head (back at start). This means we've visited all nodes exactly once.

3. Final Answer:

   Stop when current.next points to head. -> Option B

4. Quick Check:

   Circular traversal: stop when next == head, not when next is None. [OK]

Hint: Circular traversal: loop until current.next == head. Never check for None. [OK]

Common Mistakes:

• Checking for None which never occurs in circular lists
• Using a fixed count which breaks for unknown lengths
• Stopping at tail instead of checking if next == head

5. When inserting a new node at the beginning of a circular singly linked list (given head and tail), what pointer update is critical?

hard

A. Update new_node.next to None

B. Update tail.next to point to new_node (the new head)

C. Update head.next to point to new_node

D. Set tail to new_node

Solution

1. Step 1: Current state

   tail.next → head (circular link). new_node will become the new head.

2. Step 2: Required updates

   new_node.next = old_head. tail.next = new_node (maintains circular link to new head). Update head = new_node.

3. Step 3: Critical update

   tail.next must point to new_node - otherwise the circular link points to the old head, breaking the list.

4. Final Answer:

   Update tail.next to new_node to maintain the circular link. -> Option B

5. Quick Check:

   Insert at front: new.next=old_head, tail.next=new (critical!), head=new. [OK]

Hint: Insert at front: tail.next must update to new head - the circular link must be maintained. [OK]

Common Mistakes:

• Forgetting to update tail.next, breaking the circular structure
• Setting tail to new_node (tail should stay at end)
• Setting new_node.next to None instead of old head