888465. Insertion at specific position in circular linked list
Here is a detailed Markdown blog post for the problem "Insertion at specific position in circular linked list":
Insertion at specific position in circular linked list
Summary
In this problem, we are given a circular linked list and asked to insert a new node at a specific position. The key concept involved is handling the circular nature of the linked list, ensuring that the insertion does not break the circular property.
Detailed Explanation
To solve this problem, we will first need to locate the node at the specified position. Since it's a circular linked list, we can't simply traverse the list from the beginning as we would with a regular linked list. Instead, we'll use a combination of traversing and updating our current node.
Here's a step-by-step breakdown of the solution:
- Initialize three pointers:
current,new_node, andprev. Setcurrentto the head of the circular linked list,new_nodeto the new node we want to insert, andprevto null. - Traverse the circular linked list until we reach the node at the specified position. Keep track of the previous node in
prev. - Update
prev'snextpointer to point tonew_node, effectively insertingnew_nodebetweenprevandcurrent. - If the new node is the last one inserted, update its
nextpointer to the head of the circular linked list (i.e., the current node) to maintain the circular property.
Time complexity: O(n), where n is the number of nodes in the circular linked list. Space complexity: O(1), as we only use a constant amount of space to store our pointers and temporary variables.
Here's an ASCII art diagram to help illustrate the process:
+--------+
| A |
+--------+-> B
| B |
+--------+-> C
| C |
+--------+-> ...
| ... |
+--------+-> A (head of circular linked list)
Optimized Solutions
Java
public Node insertAtPosition(Node head, int position) {
if (position == 0) {
// handle edge case where new node should be the head
return new_node;
}
Node current = head;
Node prev = null;
for (int i = 1; i < position; i++) {
prev = current;
current = current.next;
}
Node new_node = // create new node
prev.next = new_node;
new_node.next = current;
return head;
}