Intersecting linked lists are two separate linked lists which share a common node or sequence of nodes. This occurs when two initially distinct lists converge at a specific point, creating a Y-shaped structure. Intersecting linked lists are often used to model real-world scenarios such as network topologies or genealogical trees.
Below is how two intersecting linked lists looks like:
Given two singly linked lists that may or may not intersect, determine if there is a point of intersection. If an intersection exists, find and return the node at which the two lists intersect.
Let’s illustrate this with examples:
Intersecting Lists: In this case, the intersecting node is the one with value 4.
Non-intersecting Lists: In this case, there is no point of intersection.
Constraints and assumptions:
The lists are singly linked, meaning each node only has a reference to the next node.
There are no cycles in either of the lists.
If an intersection exists, it is guaranteed that the nodes after the intersecting node are identical for both lists.
The intersecting node is compared by reference, not by value. Two nodes are considered intersecting only if they are the same node by reference.
Approaches to Solve the Problem
Brute Force Approach
The brute force approach is the most straightforward method to solve this problem. It involves comparing each node of one list with every node of the other list to find a common node.
Steps of the algorithm:
Start with the head of the first linked list.
For each node in the first list:
Traverse the entire second list.
Compare the current node of the first list with each node of the second list during the iteration process.
If a match is found, return this node as the intersection point.
If no match is found after comparing all nodes, return null (no intersection).
Time and Space Complexity:
Time Complexity: O(m * n), where m and n are the lengths of the two lists. In the worst case, we need to compare every node of the first list with every node of the second list.
Space Complexity: O(1), as we only use a constant amount of extra space regardless of the input size.
Below is the implementation of this approach:
# Definition of the Node class for the linked list
class Node:
def __init__(self, data):
self.data = data
self.next = None
# Helper function to create a linked list
def create_linked_list(values):
if not values:
return None
head = Node(values[0])
current = head
for value in values[1:]:
current.next = Node(value)
current = current.next
return head
# Helper function to print the linked list
def print_linked_list(head):
current = head
while current:
print(current.data, end=" -> ")
current = current.next
print("None")
def find_intersection(head1, head2):
# Start with the head of the first linked list
current1 = head1
# For each node in the first list
while current1:
# Traverse the entire second list
current2 = head2
# Compare the current node of the first list
# with each node of the second list
while current2:
# If a match is found,
# return this node as the intersection point
if current1 == current2:
return current1
current2 = current2.next
current1 = current1.next
# If no match is found after comparing all nodes,
# return None (no intersection)
return None
# Create two intersecting linked lists for testing
list1 = create_linked_list([1, 2, 3, 4, 5])
list2 = create_linked_list([9, 8])
# Create the intersection point
intersection_node = list1.next.next # Node with value 3
list2.next.next = intersection_node
print("List 1:")
print_linked_list(list1)
print("List 2:")
print_linked_list(list2)
# Find the intersection point
intersection = find_intersection(list1, list2)
if intersection:
print(f"Intersection found at node with value: {intersection.data}")
else:
print("No intersection found")
#include <iostream>
#include <vector>
using namespace std;
// Definition of the Node class for the linked list
class Node {
public:
int data;
Node* next;
Node(int value) : data(value), next(nullptr) {}
};
// Helper function to create a linked list
Node* create_linked_list(const vector<int>& values) {
if (values.empty()) {
return nullptr;
}
Node* head = new Node(values[0]);
Node* current = head;
for (size_t i = 1; i < values.size(); i++) {
current->next = new Node(values[i]);
current = current->next;
}
return head;
}
// Helper function to print the linked list
void print_linked_list(Node* head) {
Node* current = head;
while (current) {
cout << current->data << " -> ";
current = current->next;
}
cout << "nullptr" << endl;
}
Node* find_intersection(Node* head1, Node* head2) {
// Start with the head of the first linked list
Node* current1 = head1;
// For each node in the first list
while (current1) {
// Traverse the entire second list
Node* current2 = head2;
// Compare the current node of the first list
// with each node of the second list
while (current2) {
// If a match is found,
// return this node as the intersection point
if (current1 == current2) {
return current1;
}
current2 = current2->next;
}
current1 = current1->next;
}
// If no match is found after comparing all nodes,
// return nullptr (no intersection)
return nullptr;
}
int main() {
// Create two intersecting linked lists for testing
Node* list1 = create_linked_list({1, 2, 3, 4, 5});
Node* list2 = create_linked_list({9, 8});
// Create the intersection point
Node* intersection_node = list1->next->next; // Node with value 3
list2->next->next = intersection_node;
cout << "List 1:" << endl;
print_linked_list(list1);
cout << "List 2:" << endl;
print_linked_list(list2);
// Find the intersection point
Node* intersection = find_intersection(list1, list2);
if (intersection) {
cout << "Intersection found at node with value: " << intersection->data << endl;
} else {
cout << "No intersection found" << endl;
}
// Free the memory allocated for the linked lists
// Be carefull here as you need to make sure
// Not to free the common segment twice
return 0;
}
// Definition of the Node class for the linked list
class Node {
constructor(data) {
this.data = data;
this.next = null;
}
}
// Helper function to create a linked list
function createLinkedList(values) {
if (!values.length) {
return null;
}
let head = new Node(values[0]);
let current = head;
for (let i = 1; i < values.length; i++) {
current.next = new Node(values[i]);
current = current.next;
}
return head;
}
// Helper function to print the linked list
function printLinkedList(head) {
let current = head;
let result = '';
while (current) {
result += current.data + ' -> ';
current = current.next;
}
console.log(result + 'None');
}
function findIntersection(head1, head2) {
// Start with the head of the first linked list
let current1 = head1;
// For each node in the first list
while (current1) {
// Traverse the entire second list
let current2 = head2;
// Compare the current node of the first list
// with each node of the second list
while (current2) {
// If a match is found,
// return this node as the intersection point
if (current1 === current2) {
return current1;
}
current2 = current2.next;
}
current1 = current1.next;
}
// If no match is found after comparing all nodes,
// return null (no intersection)
return null;
}
// Create two intersecting linked lists for testing
let list1 = createLinkedList([1, 2, 3, 4, 5]);
let list2 = createLinkedList([9, 8]);
// Create the intersection point
let intersectionNode = list1.next.next; // Node with value 3
list2.next.next = intersectionNode;
console.log("List 1:");
printLinkedList(list1);
console.log("List 2:");
printLinkedList(list2);
// Find the intersection point
let intersection = findIntersection(list1, list2);
if (intersection) {
console.log(`Intersection found at node with value: ${intersection.data}`);
} else {
console.log("No intersection found");
}
public class Main {
// Definition of the Node class for the linked list
static class Node {
int data;
Node next;
Node(int data) {
this.data = data;
this.next = null;
}
}
// Helper function to create a linked list
static Node createLinkedList(int[] values) {
if (values.length == 0) {
return null;
}
Node head = new Node(values[0]);
Node current = head;
for (int i = 1; i < values.length; i++) {
current.next = new Node(values[i]);
current = current.next;
}
return head;
}
// Helper function to print the linked list
static void printLinkedList(Node head) {
Node current = head;
while (current != null) {
System.out.print(current.data + " -> ");
current = current.next;
}
System.out.println("null");
}
static Node findIntersection(Node head1, Node head2) {
// Start with the head of the first linked list
Node current1 = head1;
// For each node in the first list
while (current1 != null) {
// Traverse the entire second list
Node current2 = head2;
// Compare the current node of the first list
// with each node of the second list
while (current2 != null) {
// If a match is found,
// return this node as the intersection point
if (current1 == current2) {
return current1;
}
current2 = current2.next;
}
current1 = current1.next;
}
// If no match is found after comparing all nodes,
// return null (no intersection)
return null;
}
public static void main(String[] args) {
// Create two intersecting linked lists for testing
Node list1 = createLinkedList(new int[]{1, 2, 3, 4, 5});
Node list2 = createLinkedList(new int[]{9, 8});
// Create the intersection point
Node intersectionNode = list1.next.next; // Node with value 3
list2.next.next = intersectionNode;
System.out.println("List 1:");
printLinkedList(list1);
System.out.println("List 2:");
printLinkedList(list2);
// Find the intersection point
Node intersection = findIntersection(list1, list2);
if (intersection != null) {
System.out.println("Intersection found at node with value: " + intersection.data);
} else {
System.out.println("No intersection found");
}
}
}
Drawbacks of this approach:
Inefficient for large lists due to the quadratic time complexity.
Does not take advantage of the list structure or the problem constraints.
May time out for very long lists in interview or coding competition scenarios.
While this approach is easy to understand and implement, it’s not optimal for solving the problem efficiently, especially for large inputs. More sophisticated approaches can significantly improve the time complexity.
Hash Table Approach
The Hash Table approach offers a more efficient solution compared to the brute force method. This technique leverages a hash table to store nodes from one list and then checks for their presence while traversing the other list.
Steps of the algorithm:
Create an empty hash set.
Traverse the first linked list and add each node to the hash set.
Traverse the second linked list:
For each node, check if it’s already in the hash set.
If a node is found in the hash set, return it as the intersection point.
If no intersection is found after traversing both lists, return null.
Time and Space Complexity:
Time Complexity: O(m + n), where m and n are the lengths of the two lists. We traverse each list once.
Space Complexity: O(m), where m is the length of the first list. In the worst case, we store all nodes of the first list in the hash set.
Below is the implementation of this approach:
class Node:
def __init__(self, data):
self.data = data
self.next = None
# Helper function to create a linked list
def create_linked_list(values):
if not values:
return None
head = Node(values[0])
current = head
for value in values[1:]:
current.next = Node(value)
current = current.next
return head
# Helper function to print a linked list
def print_linked_list(head):
current = head
while current:
print(current.data, end=" -> ")
current = current.next
print("None")
def find_intersection(list1, list2):
# Step 1: Create an empty hash set
node_set = set()
# Step 2: Traverse the first linked list
# and add each node to the hash set
current = list1
while current:
node_set.add(current)
current = current.next
# Step 3: Traverse the second linked list
current = list2
while current:
# Check if the current node is already in the hash set
if current in node_set:
# If found, return it as the intersection point
return current
current = current.next
# Step 4: If no intersection is found, return None
return None
# Create two intersecting linked lists for testing
list1 = create_linked_list([1, 2, 3, 4, 5])
list2 = create_linked_list([9, 8])
# Create the intersection point
intersection_node = list1.next.next # Node with value 3
list2.next.next = intersection_node
print("List 1:")
print_linked_list(list1)
print("List 2:")
print_linked_list(list2)
# Find the intersection point
intersection = find_intersection(list1, list2)
if intersection:
print(f"Intersection found at node with value: {intersection.data}")
else:
print("No intersection found")
#include <iostream>
#include <unordered_set>
#include <vector>
using namespace std;
// Definition for singly-linked list node
class Node {
public:
int data;
Node* next;
Node(int val) : data(val), next(nullptr) {}
};
// Helper function to create a linked list
Node* create_linked_list(const vector<int>& values) {
if (values.empty()) {
return nullptr;
}
Node* head = new Node(values[0]);
Node* current = head;
for (size_t i = 1; i < values.size(); i++) {
current->next = new Node(values[i]);
current = current->next;
}
return head;
}
// Helper function to print a linked list
void print_linked_list(Node* head) {
Node* current = head;
while (current) {
cout << current->data << " -> ";
current = current->next;
}
cout << "nullptr" << endl;
}
Node* find_intersection(Node* list1, Node* list2) {
// Step 1: Create an empty hash set
unordered_set<Node*> node_set;
// Step 2: Traverse the first linked list
// and add each node to the hash set
Node* current = list1;
while (current) {
node_set.insert(current);
current = current->next;
}
// Step 3: Traverse the second linked list
current = list2;
while (current) {
// Check if the current node is already in the hash set
if (node_set.find(current) != node_set.end()) {
// If found, return it as the intersection point
return current;
}
current = current->next;
}
// Step 4: If no intersection is found, return nullptr
return nullptr;
}
int main() {
// Create two intersecting linked lists for testing
Node* list1 = create_linked_list({1, 2, 3, 4, 5});
Node* list2 = create_linked_list({9, 8});
// Create the intersection point
Node* intersection_node = list1->next->next; // Node with value 3
list2->next->next = intersection_node;
cout << "List 1:" << endl;
print_linked_list(list1);
cout << "List 2:" << endl;
print_linked_list(list2);
// Find the intersection point
Node* intersection = find_intersection(list1, list2);
if (intersection) {
cout << "Intersection found at node with value: " << intersection->data << endl;
} else {
cout << "No intersection found" << endl;
}
// Free memory consumed by all the nodes
// Be careful not to free the common nodes memory twice
return 0;
}
// Node class definition
class Node {
constructor(data) {
this.data = data;
this.next = null;
}
}
// Helper function to create a linked list
function createLinkedList(values) {
if (!values.length) {
return null;
}
let head = new Node(values[0]);
let current = head;
for (let i = 1; i < values.length; i++) {
current.next = new Node(values[i]);
current = current.next;
}
return head;
}
// Helper function to print a linked list
function printLinkedList(head) {
let current = head;
let result = '';
while (current) {
result += current.data + ' -> ';
current = current.next;
}
result += 'None';
console.log(result);
}
function findIntersection(list1, list2) {
// Step 1: Create an empty Set
const nodeSet = new Set();
// Step 2: Traverse the first linked list
// and add each node to the Set
let current = list1;
while (current) {
nodeSet.add(current);
current = current.next;
}
// Step 3: Traverse the second linked list
current = list2;
while (current) {
// Check if the current node is already in the Set
if (nodeSet.has(current)) {
// If found, return it as the intersection point
return current;
}
current = current.next;
}
// Step 4: If no intersection is found, return null
return null;
}
// Create two intersecting linked lists for testing
const list1 = createLinkedList([1, 2, 3, 4, 5]);
const list2 = createLinkedList([9, 8]);
// Create the intersection point
const intersectionNode = list1.next.next; // Node with value 3
list2.next.next = intersectionNode;
console.log("List 1:");
printLinkedList(list1);
console.log("List 2:");
printLinkedList(list2);
// Find the intersection point
const intersection = findIntersection(list1, list2);
if (intersection) {
console.log("Intersection found at node with value: " + intersection.data);
} else {
console.log("No intersection found");
}
import java.util.HashSet;
import java.util.Set;
public class Main {
static class Node {
int data;
Node next;
Node(int data) {
this.data = data;
this.next = null;
}
}
// Helper function to create a linked list
static Node createLinkedList(int[] values) {
if (values == null || values.length == 0) {
return null;
}
Node head = new Node(values[0]);
Node current = head;
for (int i = 1; i < values.length; i++) {
current.next = new Node(values[i]);
current = current.next;
}
return head;
}
// Helper function to print a linked list
static void printLinkedList(Node head) {
Node current = head;
while (current != null) {
System.out.print(current.data + " -> ");
current = current.next;
}
System.out.println("null");
}
static Node findIntersection(Node list1, Node list2) {
// Step 1: Create an empty hash set
Set<Node> nodeSet = new HashSet<>();
// Step 2: Traverse the first linked list
// and add each node to the hash set
Node current = list1;
while (current != null) {
nodeSet.add(current);
current = current.next;
}
// Step 3: Traverse the second linked list
current = list2;
while (current != null) {
// Check if the current node is already in the hash set
if (nodeSet.contains(current)) {
// If found, return it as the intersection point
return current;
}
current = current.next;
}
// Step 4: If no intersection is found, return null
return null;
}
public static void main(String[] args) {
// Create two intersecting linked lists for testing
Node list1 = createLinkedList(new int[]{1, 2, 3, 4, 5});
Node list2 = createLinkedList(new int[]{9, 8});
// Create the intersection point
Node intersectionNode = list1.next.next; // Node with value 3
list2.next.next = intersectionNode;
System.out.println("List 1:");
printLinkedList(list1);
System.out.println("List 2:");
printLinkedList(list2);
// Find the intersection point
Node intersection = findIntersection(list1, list2);
if (intersection != null) {
System.out.println("Intersection found at node with value: " + intersection.data);
} else {
System.out.println("No intersection found");
}
}
}
Drawbacks of this approach:
Requires additional space proportional to the size of the first list.
The constant factor for time complexity is higher due to hash computations.
If the lists are very large, the space overhead might be significant.
While this approach significantly improves the time complexity compared to the brute force method, it does so at the cost of increased space usage. For scenarios where memory is a constraint, or when dealing with extremely large lists, this trade-off might not be ideal. However, for most practical purposes, this approach offers a good balance between time efficiency and implementation simplicity.
Two-Pointer Approach (Optimal Solution)
The Two-Pointer approach provides an elegant and optimal solution to the problem of finding the intersection point of two linked lists. This method uses constant extra space and runs in linear time, making it highly efficient.
Steps of the algorithm:
Initialize two pointers, pointerA and pointerB, to the heads of the first and second lists respectively.
Traverse both lists simultaneously:
Move pointerA one step forward in the first list.
Move pointerB one step forward in the second list.
If either pointer reaches the end of its list, redirect it to the head of the other list.
Continue this process until:
The pointers meet at a node (intersection found), or
Both pointers become null (no intersection).
Return the intersection node or null.
Time and Space Complexity:
Time Complexity: O(m + n), where m and n are the lengths of the two lists. In the worst case, we traverse each list twice.
Space Complexity: O(1), as we only use two pointers regardless of the input size.
Below is the implementation of this approach:
# Definition of the Node class for the linked list
class Node:
def __init__(self, data):
self.data = data
self.next = None
# Helper function to create a linked list
def create_linked_list(values):
if not values:
return None
head = Node(values[0])
current = head
for value in values[1:]:
current.next = Node(value)
current = current.next
return head
# Helper function to print the linked list
def print_linked_list(head):
current = head
while current:
print(current.data, end=" -> ")
current = current.next
print("None")
def find_intersection(headA, headB):
# Step 1: Initialize two pointers
pointerA = headA
pointerB = headB
# Steps 2-4: Traverse both lists
while pointerA != pointerB:
# Move pointerA one step forward
if pointerA:
pointerA = pointerA.next
else:
pointerA = headB
# Move pointerB one step forward
if pointerB:
pointerB = pointerB.next
else:
pointerB = headA
# Step 5: Return the intersection node or None
return pointerA
# Example run
if __name__ == "__main__":
# Create the intersecting part of the lists
intersection = create_linked_list([3, 4, 5])
# Create the first list: 1 -> 2 -> 3 -> 4 -> 5
listA = create_linked_list([1, 2])
listA.next.next = intersection
# Create the second list: 8 -> 9 -> 3 -> 4 -> 5
listB = create_linked_list([8, 9])
listB.next.next = intersection
print("List A:")
print_linked_list(listA)
print("List B:")
print_linked_list(listB)
# Find the intersection
intersection_node = find_intersection(listA, listB)
if intersection_node:
print(f"Intersection found at node with value: {intersection_node.data}")
else:
print("No intersection found")
#include <iostream>
#include <vector>
using namespace std;
// Definition of the Node class for the linked list
class Node {
public:
int data;
Node* next;
Node(int val) : data(val), next(nullptr) {}
};
// Helper function to create a linked list
Node* create_linked_list(vector<int> values) {
if (values.empty()) {
return nullptr;
}
Node* head = new Node(values[0]);
Node* current = head;
for (int i = 1; i < values.size(); i++) {
current->next = new Node(values[i]);
current = current->next;
}
return head;
}
// Helper function to print the linked list
void print_linked_list(Node* head) {
Node* current = head;
while (current) {
cout << current->data << " -> ";
current = current->next;
}
cout << "None" << endl;
}
Node* find_intersection(Node* headA, Node* headB) {
// Step 1: Initialize two pointers
Node* pointerA = headA;
Node* pointerB = headB;
// Steps 2-4: Traverse both lists
while (pointerA != pointerB) {
// Move pointerA one step forward
if (pointerA) {
pointerA = pointerA->next;
} else {
pointerA = headB;
}
// Move pointerB one step forward
if (pointerB) {
pointerB = pointerB->next;
} else {
pointerB = headA;
}
}
// Step 5: Return the intersection node or nullptr
return pointerA;
}
int main() {
// Create the intersecting part of the lists
Node* intersection = create_linked_list({3, 4, 5});
// Create the first list: 1 -> 2 -> 3 -> 4 -> 5
Node* listA = create_linked_list({1, 2});
listA->next->next = intersection;
// Create the second list: 8 -> 9 -> 3 -> 4 -> 5
Node* listB = create_linked_list({8, 9});
listB->next->next = intersection;
cout << "List A:" << endl;
print_linked_list(listA);
cout << "List B:" << endl;
print_linked_list(listB);
// Find the intersection
Node* intersection_node = find_intersection(listA, listB);
if (intersection_node) {
cout << "Intersection found at node with value: " << intersection_node->data << endl;
} else {
cout << "No intersection found" << endl;
}
// Free the allocated memory
// Be careful not to delete the common segment twice
return 0;
}
// Definition of the Node class for the linked list
class Node {
constructor(data) {
this.data = data;
this.next = null;
}
}
// Helper function to create a linked list
function createLinkedList(values) {
if (!values.length) {
return null;
}
let head = new Node(values[0]);
let current = head;
for (let i = 1; i < values.length; i++) {
current.next = new Node(values[i]);
current = current.next;
}
return head;
}
// Helper function to print the linked list
function printLinkedList(head) {
let current = head;
let result = '';
while (current) {
result += current.data + ' -> ';
current = current.next;
}
console.log(result + 'None');
}
function findIntersection(headA, headB) {
// Step 1: Initialize two pointers
let pointerA = headA;
let pointerB = headB;
// Steps 2-4: Traverse both lists
while (pointerA !== pointerB) {
// Move pointerA one step forward
if (pointerA) {
pointerA = pointerA.next;
} else {
pointerA = headB;
}
// Move pointerB one step forward
if (pointerB) {
pointerB = pointerB.next;
} else {
pointerB = headA;
}
}
// Step 5: Return the intersection node or null
return pointerA;
}
// Example run
function main() {
// Create the intersecting part of the lists
let intersection = createLinkedList([3, 4, 5]);
// Create the first list: 1 -> 2 -> 3 -> 4 -> 5
let listA = createLinkedList([1, 2]);
listA.next.next = intersection;
// Create the second list: 8 -> 9 -> 3 -> 4 -> 5
let listB = createLinkedList([8, 9]);
listB.next.next = intersection;
console.log("List A:");
printLinkedList(listA);
console.log("List B:");
printLinkedList(listB);
// Find the intersection
let intersectionNode = findIntersection(listA, listB);
if (intersectionNode) {
console.log("Intersection found at node with value: " + intersectionNode.data);
} else {
console.log("No intersection found");
}
}
main();
public class Main {
// Definition of the Node class for the linked list
static class Node {
int data;
Node next;
Node(int data) {
this.data = data;
this.next = null;
}
}
// Helper function to create a linked list
static Node createLinkedList(int[] values) {
if (values == null || values.length == 0) {
return null;
}
Node head = new Node(values[0]);
Node current = head;
for (int i = 1; i < values.length; i++) {
current.next = new Node(values[i]);
current = current.next;
}
return head;
}
// Helper function to print the linked list
static void printLinkedList(Node head) {
Node current = head;
while (current != null) {
System.out.print(current.data + " -> ");
current = current.next;
}
System.out.println("null");
}
static Node findIntersection(Node headA, Node headB) {
// Step 1: Initialize two pointers
Node pointerA = headA;
Node pointerB = headB;
// Steps 2-4: Traverse both lists
while (pointerA != pointerB) {
// Move pointerA one step forward
if (pointerA != null) {
pointerA = pointerA.next;
} else {
pointerA = headB;
}
// Move pointerB one step forward
if (pointerB != null) {
pointerB = pointerB.next;
} else {
pointerB = headA;
}
}
// Step 5: Return the intersection node or null
return pointerA;
}
public static void main(String[] args) {
// Create the intersecting part of the lists
Node intersection = createLinkedList(new int[]{3, 4, 5});
// Create the first list: 1 -> 2 -> 3 -> 4 -> 5
Node listA = createLinkedList(new int[]{1, 2});
listA.next.next = intersection;
// Create the second list: 8 -> 9 -> 3 -> 4 -> 5
Node listB = createLinkedList(new int[]{8, 9});
listB.next.next = intersection;
System.out.println("List A:");
printLinkedList(listA);
System.out.println("List B:");
printLinkedList(listB);
// Find the intersection
Node intersectionNode = findIntersection(listA, listB);
if (intersectionNode != null) {
System.out.println("Intersection found at node with value: " + intersectionNode.data);
} else {
System.out.println("No intersection found");
}
}
}
Why this approach works:
The key insight behind this approach is that by switching the pointers to the head of the other list when they reach the end, we effectively “align” the two lists. If there’s an intersection, both pointers will meet at the intersecting node after traversing the same total distance.
Consider two lists of lengths a + c and b + c, where c is the length of the common part:
Pointer A travels: a + c + b
Pointer B travels: b + c + a
By the time both pointers have switched lists once, they will have traveled the same distance and will meet at the intersection point. If there’s no intersection, both pointers will reach the end (null) simultaneously.
This approach elegantly handles lists of different lengths and requires no additional data structures, making it both space-efficient and easy to implement.
Applications
Finding intersecting nodes in linked lists has several practical applications and is related to various real-world scenarios:
Network Topology Analysis: In computer networks, intersecting linked lists can represent converging network paths or shared resources. Identifying intersection points helps in optimizing routing algorithms and detecting bottlenecks.
Version Control Systems: Git and other VCS use linked list-like structures to represent commit histories. Finding the intersection point can help identify where two branches diverged or merged.
File Systems: In file systems with hard links, multiple directory entries can point to the same inode. Detecting intersections can help in identifying shared files and optimizing storage.
Finding the Middle of a Linked List: This problem employs a fast and slow pointer technique, conceptually similar to the two-pointer approach used in finding intersections.
Intersection of Multiple Linked Lists: An extended version of the problem where the task is to find a common node among more than two lists, increasing complexity.
Intersection with Cycles: A more intricate variation where one or both lists may contain cycles, requiring additional considerations in the solution approach.