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Intersection of two Sorted Linked Lists

  • Difficulty Level : Medium
  • Last Updated : 24 Aug, 2021
Geek Week

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++




#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 at
the 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);
}

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 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();
}

Java




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

Python3




''' 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 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'''
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.

C#




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

Javascript




<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 iterating
var 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>
Output



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();
}
Output
 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 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++




#include <bits/stdc++.h>
using namespace std;
 
// 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 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 code
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);
 
    cout << "\n Linked list containing "
         << "common items of a & b \n ";
    printList(intersect);
 
    return 0;
}
 
// This code is contributed by shivanisinghss2110

C




#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 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;
}
Output
 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 4: Use Hashing

Java




import java.util.*;
 
// This code is contributed by ayyuce demirbas
public 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);
          }
          
         
          
          
     }
      
      
      
}
Output
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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