Question

Java

implement a method that remove all duplicates in a circularly linked list
output :
before removed duplicates(7,6,5,4,3,2,1,2,4,6)
premoved duplicates(7,5,4,3,2,1,6)

Answer 1

To remove duplicates from a circularly linked list in Java, one can traverse the list and remove duplicates using a two-pointer approach where one pointer iterates and another checks for duplicates.

Step-by-step explanation:

The question pertains to the implementation of a method in Java that removes all duplicates from a circularly linked list. The method requires traversing the list and checking if any node's value has already been encountered. If a duplicate is found, it should be removed from the list.

An example approach to achieve this would be using two pointers: one for iterating through the list (current), and another one (runner) to check all subsequent nodes for duplicates. Here's a conceptual outline of this method:

1. Traverse the list with the 'current' pointer.
2. For each node that 'current' points to, use the 'runner' pointer to look ahead for duplicates.
3. If the 'runner' finds a duplicate, remove it and continue the search until the list wraps around to the starting node.
4. Move the 'current' pointer to the next node and repeat until the whole list has been checked.
5. After implementing this method, the output for the given example would change from before removed duplicates (7,6,5,4,3,2,1,2,4,6) to removed duplicates (7,5,3,2,1,6).