Intersection of two Sorted Linked Lists

Given two lists sorted in increasing order, create and return a new list representing the intersection of the two lists. The new list should be made with its own memory — the original lists should not be changed. 
Example: 

Input: 
First linked list: 1->2->3->4->6
Second linked list be 2->4->6->8, 
Output: 2->4->6.
The elements 2, 4, 6 are common in 
both the list so they appear in the 
intersection list. 

Input: 
First linked list: 1->2->3->4->5
Second linked list be 2->3->4, 
Output: 2->3->4
The elements 2, 3, 4 are common in 
both the list so they appear in the 
intersection list.

Method 1: Using Dummy Node. 
Approach: 
The idea is to use a temporary dummy node at the start of the result list. The pointer tail always points to the last node in the result list, so new nodes can be added easily. The dummy node initially gives tail a memory space to point to. This dummy node is efficient, since it is only temporary, and it is allocated in the stack. The loop proceeds, removing one node from either ‘a’ or ‘b’ and adding it to tail. When the given lists are traversed the result is in dummy.next as the values are allocated from next node of the dummy. If both the elements are equal then remove both and insert the element to the tail. Else remove the smaller element among both the lists. 

Below is the implementation of the above approach:

C

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#include <stdio.h>
#include <stdlib.h>
 
/* Link list node */
struct Node {
    int data;
    struct Node* next;
};
 
void push(struct Node** head_ref, int new_data);
 
/*This solution uses the temporary
 dummy to build up the result list */
struct Node* sortedIntersect(
    struct Node* a,
    struct Node* b)
{
    struct Node dummy;
    struct Node* tail = &dummy;
    dummy.next = NULL;
 
    /* Once one or the other
    list runs out -- we're done */
    while (a != NULL && b != NULL) {
        if (a->data == b->data) {
            push((&tail->next), a->data);
            tail = tail->next;
            a = a->next;
            b = b->next;
        }
        /* advance the smaller list */
        else if (a->data < b->data)
            a = a->next;
        else
            b = b->next;
    }
    return (dummy.next);
}
 
/* UTILITY FUNCTIONS */
/* Function to insert a node at
the beginning of the linked list */
void push(struct Node** head_ref, int new_data)
{
    /* allocate node */
    struct Node* new_node = (struct Node*)malloc(
        sizeof(struct 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(struct Node* node)
{
    while (node != NULL) {
        printf("%d ", node->data);
        node = node->next;
    }
}
 
/* Driver program to test above functions*/
int main()
{
    /* Start with the empty lists */
    struct Node* a = NULL;
    struct Node* b = NULL;
    struct Node* intersect = NULL;
 
    /* Let us create the first sorted
     linked list to test the functions
     Created linked list will be
     1->2->3->4->5->6 */
    push(&a, 6);
    push(&a, 5);
    push(&a, 4);
    push(&a, 3);
    push(&a, 2);
    push(&a, 1);
 
    /* Let us create the second sorted linked list
   Created linked list will be 2->4->6->8 */
    push(&b, 8);
    push(&b, 6);
    push(&b, 4);
    push(&b, 2);
 
    /* Find the intersection two linked lists */
    intersect = sortedIntersect(a, b);
 
    printf("\n Linked list containing common items of a & b \n ");
    printList(intersect);
 
    getchar();
}

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Output: 

Linked list containing common items of a & b 
 2 4 6 

Complexity Analysis: 



  • Time Complexity: O(m+n) where m and n are number of nodes in first and second linked lists respectively. 
    Only one traversal of the lists are needed.
  • Auxiliary Space: O(min(m, n)). 
    The output list can store at most min(m,n) nodes .

Method 2: Using Local References. 
Approach: This solution is structurally very similar to the above, but it avoids using a dummy node Instead, it maintains a struct node** pointer, lastPtrRef, that always points to the last pointer of the result list. This solves the same case that the dummy node did — dealing with the result list when it is empty. If the list is built at its tail, either the dummy node or the struct node** “reference” strategy can be used. 

Below is the implementation of the above approach:

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#include <stdio.h>
#include <stdlib.h>
 
/* Link list node */
struct Node {
    int data;
    struct Node* next;
};
 
void push(struct Node** head_ref,
          int new_data);
 
/* This solution uses the local reference */
struct Node* sortedIntersect(
    struct Node* a,
    struct Node* b)
{
    struct Node* result = NULL;
    struct Node** lastPtrRef = &result;
 
    /* Advance comparing the first
     nodes in both lists.
    When one or the other list runs
     out, we're done. */
    while (a != NULL && b != NULL) {
        if (a->data == b->data) {
            /* found a node for the intersection */
            push(lastPtrRef, a->data);
            lastPtrRef = &((*lastPtrRef)->next);
            a = a->next;
            b = b->next;
        }
        else if (a->data < b->data)
            a = a->next; /* advance the smaller list */
        else
            b = b->next;
    }
    return (result);
}
 
/* UTILITY FUNCTIONS */
/* Function to insert a node at the
   beginging of the linked list */
void push(struct Node** head_ref,
          int new_data)
{
    /* allocate node */
    struct Node* new_node = (struct Node*)malloc(
        sizeof(struct 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(struct Node* node)
{
    while (node != NULL) {
        printf("%d ", node->data);
        node = node->next;
    }
}
 
/* Driver program to test above functions*/
int main()
{
    /* Start with the empty lists */
    struct Node* a = NULL;
    struct Node* b = NULL;
    struct Node* intersect = NULL;
 
    /* Let us create the first sorted
     linked list to test the functions
   Created linked list will be
   1->2->3->4->5->6 */
    push(&a, 6);
    push(&a, 5);
    push(&a, 4);
    push(&a, 3);
    push(&a, 2);
    push(&a, 1);
 
    /* Let us create the second sorted linked list
   Created linked list will be 2->4->6->8 */
    push(&b, 8);
    push(&b, 6);
    push(&b, 4);
    push(&b, 2);
 
    /* Find the intersection two linked lists */
    intersect = sortedIntersect(a, b);
 
    printf("\n Linked list containing common items of a & b \n ");
    printList(intersect);
 
    getchar();
}

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Output: 

Linked list containing common items of a & b 
 2 4 6 

Complexity Analysis: 

  • Time Complexity: O(m+n) where m and n are number of nodes in first and second linked lists respectively. 
    Only one traversal of the lists are needed.
  • Auxiliary Space: O(max(m, n)). 
    The output list can store at most m+n nodes.

Method 3: Recursive Solution. 
Approach: 
The recursive approach is very similar to the the above two approaches. Build a recursive function that takes two nodes and returns a linked list node. Compare the first element of both the lists. 

  • If they are similar then call the recursive function with the next node of both the lists. Create a node with the data of the current node and put the returned node from the recursive function to the next pointer of the node created. Return the node created.
  • If the values are not equal then remove the smaller node of both the lists and call the recursive function.

Below is the implementation of the above approach:

C

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#include <stdio.h>
#include <stdlib.h>
 
/* Link list node */
struct Node {
    int data;
    struct Node* next;
};
 
struct Node* sortedIntersect(
    struct Node* a,
    struct Node* b)
{
    /* base case */
    if (a == NULL || b == NULL)
        return NULL;
 
    /* If both lists are non-empty */
 
    /* advance the smaller list and call recursively */
    if (a->data < b->data)
        return sortedIntersect(a->next, b);
 
    if (a->data > b->data)
        return sortedIntersect(a, b->next);
 
    // Below lines are executed only
    // when a->data == b->data
    struct Node* temp
        = (struct Node*)malloc(
            sizeof(struct Node));
    temp->data = a->data;
 
    /* advance both lists and call recursively */
    temp->next = sortedIntersect(a->next, b->next);
    return temp;
}
 
/* UTILITY FUNCTIONS */
/* Function to insert a node at
the beginging of the linked list */
void push(struct Node** head_ref, int new_data)
{
    /* allocate node */
    struct Node* new_node
        = (struct Node*)malloc(
            sizeof(struct 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(struct Node* node)
{
    while (node != NULL) {
        printf("%d ", node->data);
        node = node->next;
    }
}
 
/* Driver program to test above functions*/
int main()
{
    /* Start with the empty lists */
    struct Node* a = NULL;
    struct Node* b = NULL;
    struct Node* intersect = NULL;
 
    /* Let us create the first sorted
      linked list to test the functions
     Created linked list will be
      1->2->3->4->5->6 */
    push(&a, 6);
    push(&a, 5);
    push(&a, 4);
    push(&a, 3);
    push(&a, 2);
    push(&a, 1);
 
    /* Let us create the second sorted linked list
     Created linked list will be 2->4->6->8 */
    push(&b, 8);
    push(&b, 6);
    push(&b, 4);
    push(&b, 2);
 
    /* Find the intersection two linked lists */
    intersect = sortedIntersect(a, b);
 
    printf("\n Linked list containing common items of a & b \n ");
    printList(intersect);
 
    return 0;
}

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Output: 

Linked list containing common items of a & b 
 2 4 6 

Complexity Analysis: 

  • Time Complexity: O(m+n) where m and n are number of nodes in first and second linked lists respectively. 
    Only one traversal of the lists are needed.
  • Auxiliary Space: O(max(m, n)). 
    The output list can store at most m+n nodes.

Please write comments if you find the above codes/algorithms incorrect, or find better ways to solve the same problem.
References: 
cslibrary.stanford.edu/105/LinkedListProblems.pdf

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