C++ Program For Finding Intersection Of Two Sorted Linked Lists - GeeksforGeeks

15 Dec, 2021
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 codeint 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 ";    printList(intersect);}
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: 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++
// C++ program to implement// the above approach#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 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 << "Linked list containing " <<             "common items of a & b ";    printList(intersect);      return 0;}// This code is contributed by shivanisinghss2110
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 refer complete article on Intersection of two Sorted Linked Lists for more details!
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