Write a function that takes a list sorted in non-decreasing order and deletes any duplicate nodes from the list. The list should only be traversed once.
For example if the linked list is 11->11->11->21->43->43->60 then removeDuplicates() should convert the list to 11->21->43->60.
Algorithm:
Traverse the list from the head (or start) node. While traversing, compare each node with its next node. If the data of the next node is the same as the current node then delete the next node. Before we delete a node, we need to store the next pointer of the node
Implementation:
Functions other than removeDuplicates() are just to create a linked list and test removeDuplicates().
Python3
# Python3 program to remove duplicate# nodes from a sorted linked list # Node class class Node: # Constructor to initialize # the node object def __init__(self, data): self.data = data self.next = Noneclass LinkedList: # Function to initialize head def __init__(self): self.head = None # Function to insert a new node # at the beginning def push(self, new_data): new_node = Node(new_data) new_node.next = self.head self.head = new_node # Given a reference to the head of a # list and a key, delete the first # occurrence of key in linked list def deleteNode(self, key): # Store head node temp = self.head # If head node itself holds the # key to be deleted if (temp is not None): if (temp.data == key): self.head = temp.next temp = None return # Search for the key to be deleted, # keep track of the previous node as # we need to change 'prev.next' while(temp is not None): if temp.data == key: break prev = temp temp = temp.next # If key was not present in # linked list if(temp == None): return # Unlink the node from linked list prev.next = temp.next temp = None # Utility function to print the # linked LinkedList def printList(self): temp = self.head while(temp): print(temp.data , end = ' ') temp = temp.next # This function removes duplicates # from a sorted list def removeDuplicates(self): temp = self.head if temp is None: return while temp.next is not None: if temp.data == temp.next.data: new = temp.next.next temp.next = None temp.next = new else: temp = temp.next return self.head# Driver Code llist = LinkedList() llist.push(20) llist.push(13) llist.push(13)llist.push(11)llist.push(11)llist.push(11)print ("Created Linked List: ")llist.printList() print()print("Linked List after removing", "duplicate elements:")llist.removeDuplicates()llist.printList() # This code is contributed by Dushyant Pathak. |
Output:
Linked list before duplicate removal 11 11 11 13 13 20 Linked list after duplicate removal 11 13 20
Time Complexity: O(n) where n is the number of nodes in the given linked list.
Auxiliary Space: O(1)
Recursive Approach :
Python3
# Python3 Program to remove duplicates# from a sorted linked list import math# Link list node class Node: def __init__(self, data): self.data = data self.next = None# The function removes duplicates # from a sorted list def removeDuplicates(head): # Pointer to store the pointer # of a node to be deleted to_free # Do nothing if the list is empty if (head == None): return # Traverse the list till last node if (head.next != None): # Compare head node with next node if (head.data == head.next.data): # The sequence of steps is important. # to_free pointer stores the next of head # pointer which is to be deleted. to_free = head.next head.next = head.next.next # free(to_free) removeDuplicates(head) # This is tricky: only advance if no deletion else: removeDuplicates(head.next) return head# UTILITY FUNCTIONS # Function to insert a node at the # beginning of the linked list def push(head_ref, new_data): # Allocate node new_node = Node(new_data) # Put in the data new_node.data = new_data # Link the old list off the # new node new_node.next = head_ref # Move the head to point to the # new node head_ref = new_node return head_ref# Function to print nodes in a given # linked list def printList(node): while (node != None): print(node.data, end = " ") node = node.next # Driver codeif __name__=='__main__': # Start with the empty list head = None # Let us create a sorted linked list # to test the functions # Created linked list will be # 11.11.11.13.13.20 head = push(head, 20) head = push(head, 13) head = push(head, 13) head = push(head, 11) head = push(head, 11) head = push(head, 11) print("Linked list before duplicate removal ", end = "") printList(head) # Remove duplicates from linked list removeDuplicates(head) print("Linked list after duplicate removal ", end = "") printList(head) # This code is contributed by Srathore |
Output:
Linked list before duplicate removal 11 11 11 13 13 20 Linked list after duplicate removal 11 13 20
Time Complexity: O(n), where n is the number of nodes in the given linked list.
Auxiliary Space: O(n), due to recursive stack where n is the number of nodes in the given linked list.
Another Approach: Create a pointer that will point towards the first occurrence of every element and another pointer temp which will iterate to every element and when the value of the previous pointer is not equal to the temp pointer, we will set the pointer of the previous pointer to the first occurrence of another node.
Below is the implementation of the above approach:
Python3
# Python3 program to remove duplicates # from a sorted linked list import math # Link list node class Node: def __init__(self, data): self.data = data self.next = None # The function removes duplicates # from the given linked list def removeDuplicates(head): # Do nothing if the list consist of # only one element or empty if (head == None and head.next == None): return # Construct a pointer # pointing towards head current = head # Initialise a while loop till the # second last node of the linkedlist while (current.next): # If the data of current and next # node is equal we will skip the # node between them if current.data == current.next.data: current.next = current.next.next # If the data of current and # next node is different move # the pointer to the next node else: current = current.next return # UTILITY FUNCTIONS # Function to insert a node at the # beginning of the linked list def push(head_ref, new_data): # Allocate node new_node = Node(new_data) # Put in the data new_node.data = new_data # Link the old list off # the new node new_node.next = head_ref # Move the head to point # to the new node head_ref = new_node return head_ref # Function to print nodes # in a given linked list def printList(node): while (node != None): print(node.data, end = " ") node = node.next # Driver code if __name__=='__main__': head = None head = push(head, 20) head = push(head, 13) head = push(head, 13) head = push(head, 11) head = push(head, 11) head = push(head, 11) print("List before removal of " "duplicates ", end = "") printList(head) removeDuplicates(head) print("List after removal of " "elements ", end = "") printList(head) # This code is contributed by MukulTomar |
Output:
List before removal of duplicates 11 11 11 13 13 20 List after removal of elements 11 13 20
Time Complexity: O(n) where n is the number of nodes in the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Please refer complete article on Remove duplicates from a sorted linked list for more details!
