C++ Program For Removing Duplicates From An Unsorted Linked List
Last Updated :
23 Jul, 2025
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Write a removeDuplicates() function that takes a list and deletes any duplicate nodes from the list. The list is not sorted.
For example if the linked list is 12->11->12->21->41->43->21 then removeDuplicates() should convert the list to 12->11->21->41->43.
METHOD 1 (Using two loops):
This is the simple way where two loops are used. Outer loop is used to pick the elements one by one and the inner loop compares the picked element with the rest of the elements.
Thanks to Gaurav Saxena for his help in writing this code.
// C++ Program to remove duplicates in an
// unsorted linked list
#include <bits/stdc++.h>
using namespace std;
// A linked list node
struct Node
{
int data;
struct Node* next;
};
// Utility function to create a
// new Node
struct Node* newNode(int data)
{
Node* temp = new Node;
temp->data = data;
temp->next = NULL;
return temp;
}
/* Function to remove duplicates from a
unsorted linked list */
void removeDuplicates(struct Node* start)
{
struct Node *ptr1, *ptr2, *dup;
ptr1 = start;
// Pick elements one by one
while (ptr1 != NULL &&
ptr1->next != NULL)
{
ptr2 = ptr1;
/* Compare the picked element with rest
of the elements */
while (ptr2->next != NULL)
{
/* If duplicate then delete it */
if (ptr1->data == ptr2->next->data)
{
// Sequence of steps is important here
dup = ptr2->next;
ptr2->next = ptr2->next->next;
delete (dup);
}
else
ptr2 = ptr2->next;
}
ptr1 = ptr1->next;
}
}
// Function to print nodes in a given
// linked list
void printList(struct Node* node)
{
while (node != NULL)
{
printf("%d ", node->data);
node = node->next;
}
}
// Driver code
int main()
{
// The constructed linked list is:
// 10->12->11->11->12->11->10
struct Node* start = newNode(10);
start->next = newNode(12);
start->next->next = newNode(11);
start->next->next->next = newNode(11);
start->next->next->next->next = newNode(12);
start->next->next->next->next->next = newNode(11);
start->next->next->next->next->next->next = newNode(10);
printf("Linked list before removing duplicates ");
printList(start);
removeDuplicates(start);
printf("Linked list after removing duplicates ");
printList(start);
return 0;
}
// C++ Program to remove duplicates in an // unsorted linked list using namespace std;// A linked list node struct Node { int data; struct Node* next;};// Utility function to create a // new Nodestruct Node* newNode(int data){ Node* temp = new Node; temp->data = data; temp->next = NULL; return temp;}/* Function to remove duplicates from a unsorted linked list */void removeDuplicates(struct Node* start){ struct Node *ptr1, *ptr2, *dup; ptr1 = start; // Pick elements one by one while (ptr1 != NULL && ptr1->next != NULL) { ptr2 = ptr1; /* Compare the picked element with rest of the elements */ while (ptr2->next != NULL) { /* If duplicate then delete it */ if (ptr1->data == ptr2->next->data) { // Sequence of steps is important here dup = ptr2->next; ptr2->next = ptr2->next->next; delete (dup); } else ptr2 = ptr2->next; } ptr1 = ptr1->next; }}// Function to print nodes in a given // linked list void printList(struct Node* node){ while (node != NULL) { printf("%d ", node->data); node = node->next; }}// Driver codeint main(){ // The constructed linked list is: // 10->12->11->11->12->11->10 struct Node* start = newNode(10); start->next = newNode(12); start->next->next = newNode(11); start->next->next->next = newNode(11); start->next->next->next->next = newNode(12); start->next->next->next->next->next = newNode(11); start->next->next->next->next->next->next = newNode(10); printf("Linked list before removing duplicates "); printList(start); removeDuplicates(start); printf("Linked list after removing duplicates "); printList(start); return 0;}Output:
Linked list before removing duplicates: 10 12 11 11 12 11 10 Linked list after removing duplicates: 10 12 11
Time Complexity: O(n^2)
Auxiliary Space: O(1)
METHOD 2 (Use Sorting):
In general, Merge Sort is the best-suited sorting algorithm for sorting linked lists efficiently.
1) Sort the elements using Merge Sort. We will soon be writing a post about sorting a linked list. O(nLogn)
2) Remove duplicates in linear time using the algorithm for removing duplicates in sorted Linked List. O(n)
Please note that this method doesn't preserve the original order of elements.
Time Complexity: O(nLogn)
METHOD 3 (Use Hashing):
We traverse the link list from head to end. For every newly encountered element, we check whether it is in the hash table: if yes, we remove it; otherwise we put it in the hash table.
// C++ Program to remove duplicates in an
// unsorted linked list
#include<bits/stdc++.h>
using namespace std;
// A linked list node
struct Node
{
int data;
struct Node *next;
};
// Utility function to create a
// new Node
struct Node *newNode(int data)
{
Node *temp = new Node;
temp->data = data;
temp->next = NULL;
return temp;
}
/* Function to remove duplicates from a
unsorted linked list */
void removeDuplicates(struct Node *start)
{
// Hash to store seen values
unordered_set<int> seen;
// Pick elements one by one
struct Node *curr = start;
struct Node *prev = NULL;
while (curr != NULL)
{
// If current value is seen before
if (seen.find(curr->data) != seen.end())
{
prev->next = curr->next;
delete (curr);
}
else
{
seen.insert(curr->data);
prev = curr;
}
curr = prev->next;
}
}
// Function to print nodes in a given
// linked list
void printList(struct Node *node)
{
while (node != NULL)
{
printf("%d ", node->data);
node = node->next;
}
}
// Driver code
int main()
{
/* The constructed linked list is:
10->12->11->11->12->11->10*/
struct Node *start = newNode(10);
start->next = newNode(12);
start->next->next = newNode(11);
start->next->next->next = newNode(11);
start->next->next->next->next = newNode(12);
start->next->next->next->next->next = newNode(11);
start->next->next->next->next->next->next = newNode(10);
printf("Linked list before removing duplicates : ");
printList(start);
removeDuplicates(start);
printf("Linked list after removing duplicates : ");
printList(start);
return 0;
}
// C++ Program to remove duplicates in an // unsorted linked listusing namespace std;// A linked list node struct Node{ int data; struct Node *next;};// Utility function to create a // new Nodestruct Node *newNode(int data){ Node *temp = new Node; temp->data = data; temp->next = NULL; return temp;}/* Function to remove duplicates from a unsorted linked list */void removeDuplicates(struct Node *start){ // Hash to store seen values unordered_set<int> seen; // Pick elements one by one struct Node *curr = start; struct Node *prev = NULL; while (curr != NULL) { // If current value is seen before if (seen.find(curr->data) != seen.end()) { prev->next = curr->next; delete (curr); } else { seen.insert(curr->data); prev = curr; } curr = prev->next; }}// Function to print nodes in a given // linked list void printList(struct Node *node){ while (node != NULL) { printf("%d ", node->data); node = node->next; }}// Driver codeint main(){ /* The constructed linked list is: 10->12->11->11->12->11->10*/ struct Node *start = newNode(10); start->next = newNode(12); start->next->next = newNode(11); start->next->next->next = newNode(11); start->next->next->next->next = newNode(12); start->next->next->next->next->next = newNode(11); start->next->next->next->next->next->next = newNode(10); printf("Linked list before removing duplicates : "); printList(start); removeDuplicates(start); printf("Linked list after removing duplicates : "); printList(start); return 0;}Output:
Linked list before removing duplicates: 10 12 11 11 12 11 10 Linked list after removing duplicates: 10 12 11
Thanks to bearwang for suggesting this method.
Time Complexity: O(n) on average (assuming that hash table access time is O(1) on average).
Please refer complete article on Remove duplicates from an unsorted linked list for more details!
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