Merge two sorted linked list without duplicates

Merge two sorted linked list of size n1 and n2. The duplicates in two linked list should be present only once in the final sorted linked list.


Input : list1: 1->1->4->5->7
        list2: 2->4->7->9
Output : 1 2 4 5 7 9 

Source: Microsoft on Campus Placement and Interview Questions

Approach: Following are the steps:

  1. Merge the two sorted linked list in sorted manner. Refer recursive approach of this post. Let the final obtained list be head.
  2. Remove duplicates from sorted linked list head.




// C++ implementation to merge two sorted linked list
// without duplicates
#include <bits/stdc++.h>
using namespace std;
// structure of a node
struct Node {
    int data;
    Node* next;
// function to get a new node
Node* getNode(int data)
    // allocate space
    Node* temp = (Node*)malloc(sizeof(Node));
    // put in data
    temp->data = data;
    temp->next = NULL;
    return temp;
// function to merge two sorted linked list
// in a sorted manner
Node* sortedMerge(struct Node* a, struct Node* b)
    Node* result = NULL;
    /* Base cases */
    if (a == NULL)
        return (b);
    else if (b == NULL)
        return (a);
    /* Pick either a or b, and recur */
    if (a->data <= b->data) {
        result = a;
        result->next = sortedMerge(a->next, b);
    else {
        result = b;
        result->next = sortedMerge(a, b->next);
    return (result);
/* 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)
    /* 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;
            current->next = next_next;
        else /* This is tricky: only advance if no deletion */
            current = current->next;
// function to merge two sorted linked list
// without duplicates
Node* sortedMergeWithoutDuplicates(Node* head1, Node* head2)
    // merge two linked list in sorted manner
    Node* head = sortedMerge(head1, head2);
    // remove duplicates from the list 'head'
    return head;
// function to print the linked list
void printList(Node* head)
    while (head != NULL) {
        cout << head->data << " ";
        head = head->next;
// Driver program to test above
int main()
    // head1: 1->1->4->5->7
    Node* head1 = getNode(1);
    head1->next = getNode(1);
    head1->next->next = getNode(4);
    head1->next->next->next = getNode(5);
    head1->next->next->next->next = getNode(7);
    // head2: 2->4->7->9
    Node* head2 = getNode(2);
    head2->next = getNode(4);
    head2->next->next = getNode(7);
    head2->next->next->next = getNode(9);
    Node* head3;
    head3 = sortedMergeWithoutDuplicates(head1, head2);
    return 0;



1 2 4 5 7 9 

Time complexity: O(n1 + n2).
Auxiliary Space: O(1).

Exercise: Get the final sorted linked list without duplicates in a single traversal of the two lists.

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