# C++ Program For Removing Duplicates From A Sorted Linked List

• Last Updated : 15 Jun, 2022

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().

## C++

 `// C++ Program to remove duplicates``// from a sorted linked list``#include ``using` `namespace` `std;` `// Link list node``class` `Node``{``    ``public``:``    ``int` `data;``    ``Node* next;``};` `// The function removes duplicates``// from a sorted list``void` `removeDuplicates(Node* head)``{``    ``// Pointer to traverse the linked list``    ``Node* current = head;` `    ``// Pointer to store the next pointer``    ``// of a node to be deleted``    ``Node* next_next;``    ` `    ``// Do nothing if the list is empty``    ``if` `(current == NULL)``    ``return``;` `    ``// Traverse the list till last node``    ``while` `(current->next != NULL)``    ``{``        ``// Compare current node with next node``        ``if` `(current->data == current->next->data)``        ``{``            ``// The sequence of steps is important``            ``next_next = current->next->next;``            ``free``(current->next);``            ``current->next = next_next;``        ``}` `        ``// This is tricky: only advance if no deletion``        ``else``        ``{``            ``current = current->next;``        ``}``    ``}``}` `// UTILITY FUNCTIONS``// Function to insert a node at the``// beginning of the linked list``void` `push(Node** head_ref, ``int` `new_data)``{``    ``// Allocate node``    ``Node* new_node = ``new` `Node();``            ` `    ``// 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;``}` `// Function to print nodes in a``// given linked list``void` `printList(Node *node)``{``    ``while` `(node != NULL)``    ``{``        ``cout << ``" "` `<< node->data;``        ``node = node->next;``    ``}``}` `// Driver code``int` `main()``{``    ``// Start with the empty list``    ``Node* head = NULL;``    ` `    ``/* Let us create a sorted linked list``       ``to test the functions. Created``       ``linked list will be``       ``11->11->11->13->13->20 */``    ``push(&head, 20);``    ``push(&head, 13);``    ``push(&head, 13);``    ``push(&head, 11);``    ``push(&head, 11);``    ``push(&head, 11);                                    ` `    ``cout << ``"Linked list before duplicate removal "``;``    ``printList(head);` `    ``// Remove duplicates from linked list``    ``removeDuplicates(head);` `    ``cout << ``"Linked list after duplicate removal "``;    ``    ``printList(head);            ``    ` `    ``return` `0;``}``// This code is contributed by rathbhupendra`

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 :

## C++

 `// C++ Program to remove duplicates``// from a sorted linked list``#include ``using` `namespace` `std;` `// Link list node``class` `Node``{``    ``public``:``    ``int` `data;``    ``Node* next;``};` `// The function removes duplicates``// from a sorted list``void` `removeDuplicates(Node* head)``{``    ``// Pointer to store the pointer``    ``// of a node to be deleted*/``    ``Node* to_free;``    ` `    ``// Do nothing if the list is empty``    ``if` `(head == NULL)``        ``return``;` `    ``// Traverse the list till last node``    ``if` `(head->next != NULL)``    ``{        ``        ``// 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);``        ``}``    ``}``}` `// UTILITY FUNCTIONS``// Function to insert a node at the``// beginning of the linked list``void` `push(Node** head_ref,``          ``int` `new_data)``{``    ``// Allocate node``    ``Node* new_node = ``new` `Node();``            ` `    ``// 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;``}` `/* Function to print nodes``   ``in a given linked list */``void` `printList(Node *node)``{``    ``while` `(node != NULL)``    ``{``        ``cout << ``" "` `<< node->data;``        ``node = node->next;``    ``}``}` `// Driver code``int` `main()``{``    ``// Start with the empty list``    ``Node* head = NULL;``    ` `    ``/* Let us create a sorted linked``       ``list to test the functions``       ``Created linked list will be``       ``11->11->11->13->13->20 */``    ``push(&head, 20);``    ``push(&head, 13);``    ``push(&head, 13);``    ``push(&head, 11);``    ``push(&head, 11);``    ``push(&head, 11);                                    ` `    ``cout <<``    ``"Linked list before duplicate removal "``;``    ``printList(head);` `    ``// Remove duplicates from linked list``    ``removeDuplicates(head);` `    ``cout <<``    ``"Linked list after duplicate removal "``;    ``    ``printList(head);            ``    ` `    ``return` `0;``}``// This code is contributed by Ashita Gupta`

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:

## C++

 `// C++ program to remove duplicates``// from a sorted linked list``#include ``using` `namespace` `std;` `// Linked list Node``struct` `Node``{``    ``int` `data;``    ``Node *next;``    ``Node(``int` `d)``    ``{``      ``data = d;``      ``next = NULL;``    ``}``};` `// Function to remove duplicates``// from the given linked list``Node *removeDuplicates(Node *head)``{ ``    ``// Two references to head temp will``    ``// iterate to the whole Linked List``    ``// prev will point towards the first``    ``// occurrence of every element``    ``Node *temp = head,*prev=head;` `    ``// Traverse list till the last node``    ``while` `(temp != NULL)``    ``{``       ``// Compare values of both pointers``       ``if``(temp->data != prev->data)``       ``{        ``           ``/* if the value of prev is not equal``              ``to the value of temp that means``              ``there are no more occurrences of``              ``the prev data-> So we can set the``              ``next of prev to the temp node->*/``           ``prev->next = temp;``           ``prev = temp;``       ``}``      ` `        ``// Set the temp to the next node``        ``temp = temp->next;``    ``}``  ` `    ``/* This is the edge case if there are more than``       ``one occurrences of the last element */``    ``if``(prev != temp)``    ``{``        ``prev->next = NULL;``    ``}``    ``return` `head;``}` `Node *push(Node *head, ``int` `new_data)``{ ``    ``/* 1 & 2: Allocate the Node &``              ``Put in the data*/``    ``Node *new_node = ``new` `Node(new_data);` `    ``/* 3. Make next of new Node as head */``          ``new_node->next = head;` `    ``/* 4. Move the head to point to new Node */``    ``head = new_node;``    ``return` `head;``}` `// Function to print linked list``void` `printList(Node *head)``{``    ``Node *temp = head;``    ``while` `(temp != NULL)``    ``{``        ``cout << temp->data << ``" "``;``        ``temp = temp->next;``    ``}``    ``cout << endl;``}` `// Driver code``int` `main()``{``    ``Node *llist = NULL;``    ``llist = push(llist,20);``    ``llist = push(llist,13);``    ``llist = push(llist,13);``    ``llist = push(llist,11);``    ``llist = push(llist,11);``    ``llist = push(llist,11);``    ``cout <<``    ``(``"List before removal of duplicates"``);``    ``printList(llist);``    ``cout <<``    ``(``"List after removal of elements"``);``    ``llist = removeDuplicates(llist);``    ``printList(llist);``}``// This code is contributed by mohit kumar 29.`

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.

Another Approach: Using Maps

The idea is to push all the values in a map and printing its keys.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Link list node``struct` `Node``{``    ``int` `data;``    ``Node* next;``    ``Node()``    ``{``        ``data = 0;``        ``next = NULL;``    ``}``};` `/* Function to insert a node at``   ``the beginning of the linked``   ``list */``void` `push(Node** head_ref,``          ``int` `new_data)``{   ``    ``// Allocate node``    ``Node* new_node = ``new` `Node();` `    ``// 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;``}` `/* Function to print nodes``   ``in a given linked list */``void` `printList(Node* node)``{``    ``while` `(node != NULL)``    ``{``        ``cout << node->data << ``" "``;``        ``node = node->next;``    ``}``}` `// Function to remove duplicates``void` `removeDuplicates(Node* head)``{``    ``unordered_map<``int``, ``bool``> track;``    ``Node* temp = head;``    ``while` `(temp)``    ``{``        ``if` `(track.find(temp->data) == track.end())``        ``{``            ``cout << temp->data << ``" "``;``        ``}``        ``track[temp->data] = ``true``;``        ``temp = temp->next;``    ``}``}` `// Driver Code``int` `main()``{``    ``Node* head = NULL;` `    ``/* Created linked list will be``       ``11->11->11->13->13->20 */``    ``push(&head, 20);``    ``push(&head, 13);``    ``push(&head, 13);``    ``push(&head, 11);``    ``push(&head, 11);``    ``push(&head, 11);` `    ``cout <<``    ``"Linked list before duplicate removal "``;``    ``printList(head);` `    ``cout <<``    ``"Linked list after duplicate removal "``;``    ``removeDuplicates(head);` `    ``return` `0;``}``// This code is contributed by yashbeersingh42`

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.

Auxiliary Space: O(n) where n is the number of nodes.

Please refer complete article on Remove duplicates from a sorted linked list for more details!

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