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Q. Program to remove duplicate elements from a circular linked list.
Explanation
In this program, we will create a circular linked list and remove duplicate nodes from the list. We will compare a node with rest of the list and check for the duplicate. If the duplicate is found, delete the duplicate node from the list.
1->2->2->4->3
In the above list, we can see, node 2 is present twice in the list. So, we will have a node current that will iterate through the list. The index will point to next node to current. Temp will be pointing to the node previous to index. When a duplicate is found, we delete it by pointing temp.next to index.next. Above list after removing duplicates:
1->2->4->3
Algorithm
- Define a Node class which represents a node in the list. It has two properties data and next which will point to the next node.
- Define another class for creating the circular linked list and it has two nodes: head and tail.
- removeDuplicate() will remove duplicate nodes from the list:
- Node current will point to head and used to traverse through the list.
- The index will point to the next node to current and temp will point to previous node to index.
- We will compare the current.data with the index.data. If the match is found, delete duplicate data by pointing temp’s next to index’s next.
- Increment index to index.next and current to current .next.
- Repeat step from c to d till all the duplicates are removed.
Solution
Python
Output:
Originals list: 1 2 3 2 2 4 List after removing duplicates: 1 2 3 4
C
Output:
Originals list: 1 2 3 2 2 4 List after removing duplicates: 1 2 3 4
JAVA
Output:
Originals list: 1 2 3 2 2 4 List after removing duplicates: 1 2 3 4
C#
Output:
Originals list: 1 2 3 2 2 4 List after removing duplicates: 1 2 3 4
PHP
Output:
Originals list: 1 2 3 2 2 4 List after removing duplicates: 1 2 3 4
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