intersection

Author: sachuverma

DSA Crack Sheet/README.md

@@ -133,6 +133,7 @@
 - [Move last element to front of a given Linked List](https://www.geeksforgeeks.org/move-last-element-to-front-of-a-given-linked-list/ "view topic")
 - [Add 1 to a number represented as linked list](https://practice.geeksforgeeks.org/problems/add-1-to-a-number-represented-as-linked-list/1# "view question") - [Cpp Solution](./solutions/Add%201%20to%20a%20number%20represented%20as%20linked%20list.cpp)
 - [Add two numbers represented by linked lists](https://practice.geeksforgeeks.org/problems/add-two-numbers-represented-by-linked-lists/1# "view question") - [Cpp Solution](./solutions/Add%20two%20numbers%20represented%20by%20linked%20lists.cpp)
+- [Intersection of two sorted Linked lists](https://practice.geeksforgeeks.org/problems/intersection-of-two-sorted-linked-lists/1# "view question") - [Cpp Solution](./solutions/Intersection%20of%20two%20sorted%20Linked%20lists.cpp)
 - []( "view question") - [Cpp Solution](./solutions/.cpp)
 
 ### Binary Tree

DSA Crack Sheet/solutions/Intersection of two sorted Linked lists.cpp

@@ -0,0 +1,56 @@
+/*
+Intersection of two sorted Linked lists
+============================================
+
+Given two lists sorted in increasing order, create 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 1:
+Input:
+L1 = 1->2->3->4->6
+L2 = 2->4->6->8
+Output: 2 4 6
+Explanation: For the given first two
+linked list, 2, 4 and 6 are the elements
+in the intersection.
+
+Example 2:
+Input:
+L1 = 10->20->40->50
+L2 = 15->40
+Output: 40
+Your Task:
+The task is to complete the function intersection() which should find the intersection of two linked list and add all the elements in intersection to the third linked list and return the head of the third linked list.
+
+Expected Time Complexity : O(n+m)
+Expected Auxilliary Space : O(n+m)
+Note: n,m are the size of the linked lists.
+
+Constraints:
+1 <= size of linked lists <= 5000
+1 <= Data in linked list nodes <= 1000
+*/
+
+Node *findIntersection(Node *head1, Node *head2)
+{
+  auto curr1 = head1, curr2 = head2;
+  Node *newHead = new Node(-1);
+  auto temp = newHead;
+
+  while (curr1 && curr2)
+  {
+    if (curr1->data == curr2->data)
+    {
+      Node *newNode = new Node(curr1->data);
+      temp->next = newNode;
+      temp = temp->next;
+      curr1 = curr1->next;
+      curr2 = curr2->next;
+    }
+    else if (curr1->data < curr2->data)
+      curr1 = curr1->next;
+    else
+      curr2 = curr2->next;
+  }
+
+  return newHead->next;
+}