Delete all odd nodes of a circular linked list
In this problem, we are dealing with a circular linked list data structure. A circular linked list is a type of linked list where the last node's next pointer points back to the first node instead of being null. The problem requires us to delete all the nodes from the circular linked list that contain odd values in their data fields.
Problem Statement and Description
We have a circular linked list containing integer values. Our task is to remove all nodes with odd values from this circular linked list while keeping the list's circular structure intact. We want to achieve this by modifying the links (pointers) between nodes.
Example
Let's understand the problem with an example. Consider the following circular linked list:
Original Circular Linked List: 3 -> 12 -> 32 -> 1 -> 14 -> 27 -> 0 -> 13 -> 89 -> 3 (back to the start)
After removing the odd nodes, the circular linked list should look like this:
Modified Circular Linked List: 12 -> 32 -> 14 -> 0 -> 12 (back to the start)
Idea to Solve
To solve this problem, we need to traverse the circular linked list while maintaining a reference to the previous node (point) and the current node. We'll iterate through the list and check if the current node's data is odd. If it is odd, we'll update the pointers to remove the current node from the list. If it's not odd, we'll continue traversing.
Pseudocode
removeOddNode():
if head is null:
return
initialize temp as head
initialize point as null
initialize auxiliary as head
while temp is not null:
if temp.data is odd:
auxiliary = temp
temp = temp.next
if auxiliary is head:
if auxiliary is tail:
// Removing the only node
tail = null
head = null
temp = null
point = null
else:
head = temp
tail.next = temp
else if auxiliary is tail:
if point is not null:
point.next = head
tail = point
else:
if point is not null:
point.next = temp
auxiliary = null
else:
point = temp
temp = temp.next
if temp is head:
temp = null
Algorithm Explanation
- Check if the linked list is empty (head is null). If it is, there's nothing to remove, so return.
- Initialize the necessary pointers:
temp
to traverse the list,point
to keep track of the previous node, andauxiliary
to assist in removing nodes. - Start iterating through the list using the
temp
pointer. - Check if the data of the current node (
temp.data
) is odd. Use bitwise AND with 1 to check the least significant bit. - If the data is odd, store the reference to the current node (
temp
) in theauxiliary
pointer and movetemp
to the next node. - Check if the
auxiliary
node is the head node. If it is, handle removing the head node scenario.- If it's the only node in the list (also tail), set both
head
andtail
to null, and clear the pointers. - If it's not the only node, update
head
and maketail.next
point to the new head.
- If it's the only node in the list (also tail), set both
- If
auxiliary
is the tail node, handle removing the tail node scenario.- If
point
(previous node) is not null, updatepoint.next
to point to the head, maintaining the circular structure. - Update
tail
to the previous node (point
).
- If
- If
auxiliary
is neither the head nor the tail, handle removing an intermediate node.- Update
point.next
to point to the next node (temp
).
- Update
- Set
auxiliary
to null to prevent accidental access. - If the current node's data is not odd, it's an even node. Update the
point
pointer totemp
and movetemp
to the next node. - Check if
temp
has reached back to the head, indicating completion of one loop through the circular list. In this case, settemp
to null to terminate the loop.
Code Solution
-
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Time Complexity
The time complexity of the removeOddNode
function is O(N), where N is the number of nodes in the
circular linked list. This is because we are traversing through all the nodes once to identify and remove odd nodes.
The operations within the loop, such as updating pointers, are constant-time operations.
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