Intersection of two Sorted Linked Lists

01 Jul, 2022
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 the 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 the 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++
C
Java
Python3
C#
Javascript
#include<bits/stdc++.h>using namespace std; /* Link list node */struct Node {    int data;    Node* next;}; void push(Node** head_ref, int new_data); /*This solution uses the temporary dummy to build up the result list */Node* sortedIntersect(Node* a, Node* b){    Node dummy;    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 atthe beginning of the linked list */void push(Node** head_ref, int new_data){    /* allocate node */    Node* new_node = (Node*)malloc(        sizeof(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 program to test above functions*/int main(){    /* Start with the empty lists */    Node* a = NULL;    Node* b = NULL;    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);     cout<<"Linked list containing common items of a & b \n";    printList(intersect);}
#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 atthe 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();}
class GFG{       // head nodes for pointing to 1st and 2nd linked lists    static Node a = null, b = null;       // dummy node for storing intersection    static Node dummy = null;       // tail node for keeping track of  // last node so that it makes easy for insertion    static Node tail = null;         // class - Node    static class Node {        int data;        Node next;         Node(int data) {            this.data = data;            next = null;        }    }         // function for printing the list    void printList(Node start) {        Node p = start;        while (p != null) {            System.out.print(p.data + " ");            p = p.next;        }        System.out.println();    }         // inserting elements into list    void push(int data) {        Node temp = new Node(data);        if(dummy == null) {            dummy = temp;            tail = temp;        }        else {            tail.next = temp;            tail = temp;        }    }         // function for finding intersection and adding it to dummy list    void sortedIntersect()    {               // pointers for iterating        Node p = a,q = b;        while(p != null  &&  q != null)        {            if(p.data == q.data)            {                // add to dummy list                push(p.data);                p = p.next;                q = q.next;            }            else if(p.data < q.data)                p = p.next;            else                q= q.next;        }    }       // Driver code    public static void main(String args[])    {        GFG list = new GFG();                 // creating first linked list        list.a = new Node(1);        list.a.next = new Node(2);        list.a.next.next = new Node(3);        list.a.next.next.next = new Node(4);        list.a.next.next.next.next = new Node(6);         // creating second linked list        list.b = new Node(2);        list.b.next = new Node(4);        list.b.next.next = new Node(6);        list.b.next.next.next = new Node(8);                 // function call for intersection        list.sortedIntersect();               // print required intersection        System.out.println("Linked list containing common items of a & b");        list.printList(dummy);    }} // This code is contributed by Likhita AVL
''' Link list node '''class Node:    def __init__(self):          self.data = 0        self.next = None     '''This solution uses the temporary dummy to build up the result list '''def sortedIntersect(a, b):    dummy = Node()    tail = dummy;    dummy.next = None;      ''' Once one or the other    list runs out -- we're done '''    while (a != None and b != None):        if (a.data == b.data):            tail.next = push((tail.next), a.data);            tail = tail.next;            a = a.next;            b = b.next;                 # advance the smaller list        elif(a.data < b.data):            a = a.next;        else:            b = b.next;       return (dummy.next); ''' UTILITY FUNCTIONS '''''' Function to insert a node atthe beginning of the linked list '''def push(head_ref, new_data):     ''' allocate node '''    new_node = 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;       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__':         ''' Start with the empty lists '''    a = None;    b = None;    intersect = None;      ''' Let us create the first sorted     linked list to test the functions     Created linked list will be     1.2.3.4.5.6 '''    a = push(a, 6);    a = push(a, 5);    a = push(a, 4);    a = push(a, 3);    a = push(a, 2);    a = push(a, 1);      ''' Let us create the second sorted linked list   Created linked list will be 2.4.6.8 '''    b = push(b, 8);    b = push(b, 6);    b = push(b, 4);    b = push(b, 2);      ''' Find the intersection two linked lists '''    intersect = sortedIntersect(a, b);      print("Linked list containing common items of a & b ");    printList(intersect); # This code is contributed by rutvik_56.
using System; public class GFG{       // dummy node for storing intersection    static Node dummy = null;       // tail node for keeping track of  // last node so that it makes easy for insertion    static Node tail = null;         // class - Node  public  class Node  {       public int data;      public  Node next;       public  Node(int data)      {            this.data = data;            next = null;        }    }        // head nodes for pointing to 1st and 2nd linked lists    Node a = null, b = null;            // function for printing the list    void printList(Node start)    {        Node p = start;        while (p != null)        {            Console.Write(p.data + " ");            p = p.next;        }        Console.WriteLine();    }         // inserting elements into list    void push(int data)    {        Node temp = new Node(data);        if(dummy == null)        {            dummy = temp;            tail = temp;        }        else        {            tail.next = temp;            tail = temp;        }    }         // function for finding intersection and adding it to dummy list    void sortedIntersect()    {               // pointers for iterating        Node p = a,q = b;        while(p != null  &&  q != null)        {            if(p.data == q.data)            {                // add to dummy list                push(p.data);                p = p.next;                q = q.next;            }            else if(p.data < q.data)                p = p.next;            else                q= q.next;        }    }       // Driver code    public static void Main(String []args)    {        GFG list = new GFG();                 // creating first linked list        list.a = new Node(1);        list.a.next = new Node(2);        list.a.next.next = new Node(3);        list.a.next.next.next = new Node(4);        list.a.next.next.next.next = new Node(6);         // creating second linked list        list.b = new Node(2);        list.b.next = new Node(4);        list.b.next.next = new Node(6);        list.b.next.next.next = new Node(8);                 // function call for intersection        list.sortedIntersect();               // print required intersection        Console.WriteLine("Linked list containing common items of a & b");        list.printList(dummy);    }} // This code is contributed by aashish1995
<script>     // head nodes for pointing to    // 1st and 2nd linked lists    var a = null, b = null;     // dummy node for storing intersection    var dummy = null;     // tail node for keeping track of    // last node so that it makes easy for insertion    var tail = null;     // class - Node     class Node {            constructor(val) {                this.data = val;                this.next = null;            }        }           // function for printing the list    function printList(start) {var p = start;        while (p != null) {            document.write(p.data + " ");            p = p.next;        }        document.write();    }     // inserting elements into list    function push(data) {var temp = new Node(data);        if (dummy == null) {            dummy = temp;            tail = temp;        } else {            tail.next = temp;            tail = temp;        }    }     // function for finding intersection and    // adding it to dummy list    function sortedIntersect() {         // pointers for iteratingvar p = a, q = b;        while (p != null && q != null) {            if (p.data == q.data) {                // add to dummy list                push(p.data);                p = p.next;                q = q.next;            } else if (p.data < q.data)                p = p.next;            else                q = q.next;        }    }     // Driver code              // creating first linked list        a = new Node(1);        a.next = new Node(2);        a.next.next = new Node(3);        a.next.next.next = new Node(4);        a.next.next.next.next = new Node(6);         // creating second linked list        b = new Node(2);        b.next = new Node(4);        b.next.next = new Node(6);        b.next.next.next = new Node(8);         // function call for intersection        sortedIntersect();         // print required intersection        document.write(        "Linked list containing common items of a & b<br/>"        );        printList(dummy); // This code is contributed by todaysgaurav </script>
Linked list containing common items of a & b 
2 4 6 
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:
C
#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   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();}
 Linked list containing common items of a & b 
 2 4 6 
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 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++
C
Java
Python3
C#
Javascript
#include <bits/stdc++.h>using namespace std; // Link list nodestruct 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 atthe 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)    {        cout << " " << node->data;        node = node->next;    }} // Driver codeint 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);     cout << "\n Linked list containing "         << "common items of a & b \n ";    printList(intersect);     return 0;} // This code is contributed by shivanisinghss2110
#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 atthe 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);     return 0;}
import java.util.*; class GFG{ // Link list nodestatic class Node{    int data;    Node next;}; static Node sortedIntersect(Node a,                             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    Node temp = new 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 atthe beginning of the linked list */static Node 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;    return head_ref;} /* Function to print nodes in   a given linked list */static void printList(Node node){    while (node != null)    {        System.out.print(" " +  node.data);        node = node.next;    }} // Driver codepublic static void main(String[] args){         /* Start with the empty lists */    Node a = null;    Node b = null;    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 */    a = push(a, 6);    a = push(a, 5);    a = push(a, 4);    a = push(a, 3);    a = push(a, 2);    a = push(a, 1);     /* Let us create the second sorted linked list     Created linked list will be 2.4.6.8 */    b = push(b, 8);    b = push(b, 6);    b = push(b, 4);    b = push(b, 2);     /* Find the intersection two linked lists */    intersect = sortedIntersect(a, b);     System.out.print("\n Linked list containing "        + "common items of a & b \n ");    printList(intersect); }} // This code is contributed by umadevi9616
# Link list nodeclass Node:    def __init__(self):        self.data = 0        self.next = None          def sortedIntersect(a, b):     # base case    if (a == None or b == None):        return None     # 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    temp = 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 beginning of the linked list def push(head_ref, new_data):     # Allocate node    new_node = 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;    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 # Start with the empty listsa = Noneb = Noneintersect = None # Let us create the first sorted linked list to test the functions Created# linked list will be 1.2.3.4.5.6         a = push(a, 6)a = push(a, 5)a = push(a, 4)a = push(a, 3)a = push(a, 2)a = push(a, 1) # Let us create the second sorted linked list Created linked list will be# 2.4.6.8         b = push(b, 8)b = push(b, 6)b = push(b, 4)b = push(b, 2) # Find the intersection two linked listsintersect = sortedIntersect(a, b) print("\n Linked list containing " + "common items of a & b");printList(intersect) # This code is contributed by Saurabh Jaiswal
using System; public class GFG {   // Link list node  public    class Node {    public    int data;    public    Node next;  };   static Node sortedIntersect(Node a, 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    Node temp = new 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 beginning of the linked list     */  static Node 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;    return head_ref;  }   /*     * Function to print nodes in a given linked list     */  static void printList(Node node) {    while (node != null) {      Console.Write(" " + node.data);      node = node.next;    }  }   // Driver code  public static void Main(String[] args) {     /* Start with the empty lists */    Node a = null;    Node b = null;    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         */    a = push(a, 6);    a = push(a, 5);    a = push(a, 4);    a = push(a, 3);    a = push(a, 2);    a = push(a, 1);     /*         * Let us create the second sorted linked list Created linked list will be         * 2.4.6.8         */    b = push(b, 8);    b = push(b, 6);    b = push(b, 4);    b = push(b, 2);     /* Find the intersection two linked lists */    intersect = sortedIntersect(a, b);     Console.Write("\n Linked list containing " + "common items of a & b \n ");    printList(intersect);   }} // This code is contributed by umadevi9616
<script>    // Link list node     class Node {     constructor(){        this.data = 0;        this.next = null;}    }         function sortedIntersect(a,  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.datavar temp = new 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 beginning of the linked list     */    function push(head_ref , new_data) {         /* Allocate node */var 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;        return head_ref;    }     /*     * Function to print nodes in a given linked list     */    function printList(node) {        while (node != null) {            document.write(" " + node.data);            node = node.next;        }    }     // Driver code              /* Start with the empty lists */var a = null;var b = null;var 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         */        a = push(a, 6);        a = push(a, 5);        a = push(a, 4);        a = push(a, 3);        a = push(a, 2);        a = push(a, 1);         /*         * Let us create the second sorted linked list Created linked list will be         * 2.4.6.8         */        b = push(b, 8);        b = push(b, 6);        b = push(b, 4);        b = push(b, 2);         /* Find the intersection two linked lists */        intersect = sortedIntersect(a, b);         document.write("\n Linked list containing "        + "common items of a & b <br/> ");        printList(intersect); // This code is contributed by Rajput-Ji</script>
 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 4: Use Hashing
Java
C#
import java.util.*; // This code is contributed by ayyuce demirbaspublic class LinkedList {    Node head;     static class Node {            int data;            Node next;                                      Node(int d)  {                data = d;                 next=null;                      } }     public void printList() {         Node n= head;         while(n!=null) {             System.out.println(n.data+ " ");             n=n.next;         }     }                       public void append(int d) {                         Node n= new Node(d);                         if(head== null) {                head= new Node(d);                return;            }                                      n.next=null;            Node last= head;            while(last.next !=null) {                last=last.next;                }                last.next=n;                return;                     }                 static int[] intersection(Node tmp1, Node tmp2, int k) {            int[] res = new int[k];                         HashSet<Integer> set = new HashSet<Integer>();            while(tmp1 != null) {                                 set.add(tmp1.data);                tmp1=tmp1.next;                             }                         int cnt=0;                         while(tmp2 != null) {                if(set.contains(tmp2.data)) {                    res[cnt]=tmp2.data;                    cnt++;                }                tmp2=tmp2.next;                             }                         return res;                     }                 public static void main(String[] args) {         LinkedList ll = new LinkedList();         LinkedList ll1 = new LinkedList();                   ll.append(0);         ll.append(1);         ll.append(2);         ll.append(3);         ll.append(4);         ll.append(5);         ll.append(6);         ll.append(7);                             ll1.append(9);         ll1.append(0);         ll1.append(12);         ll1.append(3);         ll1.append(4);         ll1.append(5);         ll1.append(6);         ll1.append(7);                              int[] arr= intersection(ll.head, ll1.head,6);                     for(int i : arr) {              System.out.println(i);          }                                            }                  }
using System;using System.Collections.Generic; // This code is contributed by ayyuce demirbaspublic class List {  Node head;   public class Node {    public int data;    public Node next;     public Node(int d)    {      data = d;      next = null;    }  }   public void printList()  {    Node n = head;    while (n != null) {      Console.WriteLine(n.data + " ");      n = n.next;    }  }   public void append(int d)  {     Node n = new Node(d);     if (head == null) {      head = new Node(d);      return;    }     n.next = null;    Node last = head;    while (last.next != null) {      last = last.next;    }    last.next = n;    return;  }   static int[] intersection(Node tmp1, Node tmp2, int k)  {    int[] res = new int[k];     HashSet<int> set = new HashSet<int>();    while (tmp1 != null) {       set.Add(tmp1.data);      tmp1 = tmp1.next;    }     int cnt = 0;     while (tmp2 != null) {      if (set.Contains(tmp2.data)) {        res[cnt] = tmp2.data;        cnt++;      }      tmp2 = tmp2.next;    }     return res;  }   public static void Main(String[] args)  {    List ll = new List();    List ll1 = new List();     ll.append(0);    ll.append(1);    ll.append(2);    ll.append(3);    ll.append(4);    ll.append(5);    ll.append(6);    ll.append(7);     ll1.append(9);    ll1.append(0);    ll1.append(12);    ll1.append(3);    ll1.append(4);    ll1.append(5);    ll1.append(6);    ll1.append(7);     int[] arr = intersection(ll.head, ll1.head, 6);     foreach(int i in arr) { Console.WriteLine(i); }  }} // This code is contributed by umadevi9616
0
3
4
5
6
7
Complexity Analysis:
Time Complexity: O(n)
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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