C Exercises: Find the intersection of two singly linked lists
C Singly Linked List : Exercise-19 with Solution
Write a C program to find the intersection of two singly linked lists.
Sample Solution:
C Code:
#include<stdio.h>
#include <stdlib.h>
// Define a structure for a Node in a singly linked list
struct Node {
int data;
struct Node* next;
};
// Function to create a new node with given data
struct Node* newNode(int data) {
// Allocate memory for a new node
struct Node* node = (struct Node*) malloc(sizeof(struct Node));
// Set the data for the new node
node->data = data;
// Set the next pointer of the new node to NULL
node->next = NULL;
return node;
}
// Function to find the intersection node of two linked lists
struct Node* getIntersection(struct Node* head1, struct Node* head2) {
int count1 = 0, count2 = 0;
struct Node *curr1 = head1, *curr2 = head2;
// Count the number of nodes in each list
while (curr1) {
count1++;
curr1 = curr1->next;
}
while (curr2) {
count2++;
curr2 = curr2->next;
}
// Reset the pointers to the heads of the lists
curr1 = head1;
curr2 = head2;
// Adjust the pointers of the larger list to have the same number of nodes as the smaller list
if (count1 > count2) {
int i;
for (i = 0; i < count1 - count2; i++)
curr1 = curr1->next;
} else {
int i;
for (i = 0; i < count2 - count1; i++)
curr2 = curr2->next;
}
// Move both pointers together until they meet at the intersection
while (curr1 && curr2) {
if (curr1 == curr2)
return curr1; // Intersection found
curr1 = curr1->next;
curr2 = curr2->next;
}
// If there is no intersection, return NULL
return NULL;
}
// Function to display the elements of a linked list
void displayList(struct Node* head) {
while (head) {
printf("%d ", head->data);
head = head->next;
}
printf("\n");
}
// Main function where the execution starts
int main() {
// Creating two linked lists
struct Node* head1 = newNode(1);
head1->next = newNode(2);
head1->next->next = newNode(3);
head1->next->next->next = newNode(4);
struct Node* head2 = newNode(5);
head2->next = head1->next->next;
printf("Original lists:\n");
displayList(head1);
displayList(head2);
// Finding intersection in the first pair of lists
struct Node* intersection1 = getIntersection(head1, head2);
if (intersection1)
printf("Intersection found at %d.\n", intersection1->data);
else
printf("Intersection not found.\n");
// Creating another pair of linked lists
struct Node* head3 = newNode(1);
head3->next = newNode(2);
head3->next->next = newNode(3);
head3->next->next->next = newNode(4);
struct Node* head4 = newNode(5);
head4->next = newNode(3);
head4->next->next = newNode(4);
printf("\nOriginal lists:\n");
displayList(head3);
displayList(head4);
// Finding intersection in the second pair of lists
struct Node* intersection2 = getIntersection(head3, head4);
if (intersection2)
printf("Intersection found at %d.\n", intersection2->data);
else
printf("Intersection not found.\n");
return 0; // Indicates successful completion of the program
}
Sample Output:
Original lists: 1 2 3 4 5 3 4 Intersection found at 3. Original lists: 1 2 3 4 5 3 4 Intersection not found.
Flowchart :
C Programming Code Editor:
Previous: Copy of a singly linked list with random pointers.
Next: Nodes from the end of a singly linked list.
What is the difficulty level of this exercise?
It will be nice if you may share this link in any developer community or anywhere else, from where other developers may find this content. Thanks.
https://www.w3resource.com/c-programming-exercises/linked_list/c-linked_list-exercise-41.php
- Weekly Trends and Language Statistics
- Weekly Trends and Language Statistics