Reverse a doubly circular linked list
The problem involves reversing a circular doubly linked list by swapping the values of the nodes. A circular doubly linked list is a data structure in which each node has pointers to both its previous and next nodes, and the last node's next pointer points back to the first node, creating a loop.
Problem Statement
Given a circular doubly linked list, the task is to reverse the order of its elements using a swap of node values.
Example
Consider the circular doubly linked list: 1 <-> 2 <-> 3 <-> 4 <-> 5 <-> 6. After reversing the elements, the list will become: 6 <-> 5 <-> 4 <-> 3 <-> 2 <-> 1.
Idea to Solve the Problem
To reverse the order of elements in a circular doubly linked list using a swap of node values, we need to follow these steps:
- Get references to the first node (
front
) and the last node (tail
). - Initialize a temporary variable (
temp
) to store the value during the swap operation. - Traverse the list using two pointers (
first
andlast
), moving towards each other from the front and the tail respectively. - Swap the values of the nodes pointed to by
first
andlast
. - Continue swapping and moving the pointers until
first
andlast
meet or cross each other. - The reversal is complete when
first
andlast
cross each other.
Pseudocode
Here's the pseudocode to reverse a circular doubly linked list by swapping node values:
function reverseElement():
first = front
last = tail
while first is not NULL and last is not NULL:
if first is not last:
// Swap values of first and last nodes
temp = first.data
first.data = last.data
last.data = temp
if first.next is last:
// Stop reversal
first = NULL
last = NULL
else:
first = first.next
last = last.back
else:
// Stop reversal
first = NULL
last = NULL
Algorithm Explanation
- Get references to the first node (
front
) and the last node (tail
). - Initialize a temporary variable (
temp
) to store the value during the swap operation. - Traverse the list using two pointers (
first
andlast
), moving towards each other from the front and the tail respectively. - Swap the values of the nodes pointed to by
first
andlast
. - Continue swapping and moving the pointers until
first
andlast
meet or cross each other. - The reversal is complete when
first
andlast
cross each other.
Code Solution
-
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Time Complexity
The time complexity of reversing a circular doubly linked list using a swap of node values is O(n), where n is the number of nodes in the list. This is because we iterate through the list once to perform the swap operation.
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