Question

// Removes all copies of o from this linked list.

// You have to handle the cases where Object o may

// have zero, one or multiple copies in this list.

// If any element has been removed from this list, returns true.

// Otherwise returns false.

// Note: be careful when multiple copies of Object o are stored

// in consecutive(adjacent) list nodes.

// E.g. []->["A"]->["A"]->["B"].

// Be careful when removing all "A"s in this example.

// Note: This list may contains zero, one or multiple copies of null data elements.

public boolean removeAllCopies( Object o ) {

//code here

}

Answer 1

linked_list_data = [x for x in input().split()]

#PRINTING THE CONTENT OF LINKED list

def printList(linked_list_data):
for i in range(len(linked_list_data)):
print("{0}->".format(linked_list_data[i]),end='')
print("NULL")
  

def removeObject(object_char):
global linked_list_data
linked_list_data = list(filter((object_char).__ne__,linked_list_data))

printList(linked_list_data)
object_char = input("Enter the object you want to remove: ")
removeObject(object_char)
print()
printList(linked_list_data)

updated answer as per student requirement