Intersection Point of Two Linked Lists in DSA C++ - Build from Scratch

Intersection Point of Two Linked Lists

📖 Scenario: Imagine two train tracks that merge into one track at some point. Each track is like a linked list, and the merging point is the intersection node where both linked lists meet. We want to find this intersection point.

🎯 Goal: Build a program that creates two linked lists with a shared intersection node and finds the intersection point node using pointers.

📋 What You'll Learn

Create two linked lists with some nodes

Manually create an intersection by linking the second list to a node in the first list

Implement a function to find the intersection point of the two linked lists

Print the value of the intersection node or -1 if no intersection exists

💡 Why This Matters

🌍 Real World

Finding intersection points in linked data structures is useful in memory management, network routing, and detecting shared resources.

💼 Career

Understanding linked list intersections helps in debugging pointer-related bugs and optimizing data structure usage in software engineering roles.

1

Create two linked lists with nodes

Create a struct called Node with an int data and a Node* next. Then create two linked lists: head1 with nodes containing values 3, 6, 9, and head2 with nodes containing values 10, 15. Initialize head1 and head2 accordingly.

DSA C++

// Define Node struct
// Create head1 with nodes 3 -> 6 -> 9 -> nullptr
// Create head2 with nodes 10 -> 15 -> nullptr

Hint

Define the Node struct first. Then create nodes for each list by linking them with next.

2

Create the intersection point

Link the next of the last node of head2 (node with value 15) to the node with value 9 in head1 to create an intersection.

DSA C++

#include 

struct Node {
    int data;
    Node* next;
    Node(int val) : data(val), next(nullptr) {}
};

int main() {
    Node* head1 = new Node(3);
    head1->next = new Node(6);
    head1->next->next = new Node(9);

    Node* head2 = new Node(10);
    head2->next = new Node(15);

    // Link head2's last node to head1's node with value 9
    // Your code here

    return 0;
}

Hint

Set the next pointer of the last node in head2 to point to the node with value 9 in head1.

3

Implement function to find intersection point

Write a function Node* getIntersectionNode(Node* head1, Node* head2) that returns the intersection node of the two linked lists or nullptr if there is no intersection. Use two pointers ptr1 and ptr2 to traverse the lists and switch heads when reaching the end.

DSA C++

#include 

struct Node {
    int data;
    Node* next;
    Node(int val) : data(val), next(nullptr) {}
};

Node* getIntersectionNode(Node* head1, Node* head2) {
    // Use two pointers ptr1 and ptr2
    // Your code here
}

int main() {
    Node* head1 = new Node(3);
    head1->next = new Node(6);
    head1->next->next = new Node(9);

    Node* head2 = new Node(10);
    head2->next = new Node(15);

    head2->next->next = head1->next->next; // Intersection at node with value 9

    return 0;
}

Hint

Use two pointers starting at each head. Move them forward. When a pointer reaches the end, switch it to the other list's head. They will meet at the intersection or both become nullptr.

4

Print the intersection node data

Call getIntersectionNode(head1, head2) and print the data of the returned node. If the returned node is nullptr, print -1.

DSA C++

#include 

struct Node {
    int data;
    Node* next;
    Node(int val) : data(val), next(nullptr) {}
};

Node* getIntersectionNode(Node* head1, Node* head2) {
    Node* ptr1 = head1;
    Node* ptr2 = head2;

    while (ptr1 != ptr2) {
        ptr1 = (ptr1 == nullptr) ? head2 : ptr1->next;
        ptr2 = (ptr2 == nullptr) ? head1 : ptr2->next;
    }
    return ptr1;
}

int main() {
    Node* head1 = new Node(3);
    head1->next = new Node(6);
    head1->next->next = new Node(9);

    Node* head2 = new Node(10);
    head2->next = new Node(15);

    head2->next->next = head1->next->next; // Intersection at node with value 9

    // Call getIntersectionNode and print the data or -1 if nullptr
    // Your code here

    return 0;
}

Hint

Call the function and check if the result is not nullptr. Print the data or -1 accordingly.

Practice

1. What does the term intersection point of two linked lists mean?

easy

A. The last node of each linked list.

B. The node where the values of both linked lists are equal.

C. The first node of the longer linked list.

D. The node where both linked lists share the same reference (memory address).

Solution

1. Step 1: Understand linked list nodes

   Each node in a linked list has a value and a reference (address) to the next node.

2. Step 2: Define intersection point

   The intersection point is where two linked lists share the exact same node reference, not just the same value.

3. Final Answer:

   The node where both linked lists share the same reference (memory address). -> Option D

4. Quick Check:

   Intersection means shared node reference = A [OK]

Hint: Intersection means same node object, not just same value [OK]

Common Mistakes:

• Confusing same value with same node reference
• Thinking intersection is the last node always
• Assuming intersection means equal length lists

2. Which of the following C++ code snippets correctly declares a singly linked list node structure for intersection detection?

easy

A. struct Node { int data; Node* next; };

B. struct Node { int data; Node next; };

C. struct Node { int data; Node* next(){}; };

D. struct Node { int data; Node next(); };

Solution

1. Step 1: Understand pointer usage in C++ structs

   A linked list node must have a pointer to the next node, so Node* next; is correct.

2. Step 2: Analyze options

   struct Node { int data; Node* next; }; uses Node* next; which is correct. Options A, B, and D misuse pointer or function syntax.

3. Final Answer:

   struct Node { int data; Node* next; }; -> Option A

4. Quick Check:

   Pointer to next node = C [OK]

Hint: Use pointer (*) for next node in struct [OK]

Common Mistakes:

• Forgetting the * pointer for next node
• Using function syntax instead of pointer
• Declaring next as object instead of pointer

3. Given two linked lists intersecting at node with value 7, what will the following code print?

Node* getIntersectionNode(Node* headA, Node* headB) {
    Node* a = headA;
    Node* b = headB;
    while (a != b) {
        a = a ? a->next : headB;
        b = b ? b->next : headA;
    }
    return a;
}

// Lists:
// A: 3 -> 6 -> 9 -> 7 -> 10
// B: 1 -> 7 -> 10

Node* intersection = getIntersectionNode(headA, headB);
if (intersection) std::cout << intersection->data << std::endl;
else std::cout << -1 << std::endl;

medium

A. -1

B. 10

C. 7

D. 3

Solution

1. Step 1: Understand the algorithm

   This code uses two pointers traversing both lists; when one reaches end, it switches to the other list's head. They meet at intersection.

2. Step 2: Trace the pointers

   Both pointers meet at node with value 7, which is the intersection point.

3. Final Answer:

   7 -> Option C

4. Quick Check:

   Intersection node data = 7 [OK]

Hint: Two-pointer switch method finds intersection node [OK]

Common Mistakes:

• Assuming intersection is node with value 10
• Printing -1 when intersection exists
• Confusing node value with node reference

4. Identify the bug in this code snippet that tries to find the intersection point of two linked lists:

Node* getIntersection(Node* headA, Node* headB) {
    while (headA != headB) {
        headA = headA->next;
        headB = headB->next;
    }
    return headA;
}

medium

A. It does not handle different list lengths, causing null pointer errors.

B. It uses incorrect comparison operator for nodes.

C. It returns the wrong node after loop ends.

D. It should use recursion instead of iteration.

Solution

1. Step 1: Analyze pointer movement

   The code moves both pointers forward simultaneously without adjusting for different list lengths.

2. Step 2: Identify consequence

   If lists differ in length, one pointer may reach null before the other, causing a null pointer dereference or infinite loop.

3. Final Answer:

   It does not handle different list lengths, causing null pointer errors. -> Option A

4. Quick Check:

   Unequal lengths cause errors = D [OK]

Hint: Adjust pointers for length difference before comparing [OK]

Common Mistakes:

• Ignoring list length difference
• Dereferencing null pointers
• Assuming simultaneous traversal always works

5. You have two singly linked lists that intersect. List A has length 8, List B has length 5. Which approach correctly finds the intersection point efficiently?

hard

A. Traverse both lists fully and store nodes in hash sets, then find first common node.

B. Calculate length difference, advance longer list by difference, then move both pointers together.

C. Reverse both lists and compare nodes from start until they differ.

D. Compare node values one by one from heads until a matching value is found.

Solution

1. Step 1: Understand length difference method

   Calculate difference (8 - 5 = 3), advance longer list by 3 nodes to align both lists.

2. Step 2: Traverse both lists together

   Move both pointers one step at a time; the first node where they match is the intersection.

3. Final Answer:

   Calculate length difference, advance longer list by difference, then move both pointers together. -> Option B

4. Quick Check:

   Align by length difference then traverse = B [OK]

Hint: Advance longer list by length difference before simultaneous traversal [OK]

Common Mistakes:

• Comparing node values instead of references
• Reversing lists unnecessarily
• Using extra space when not needed