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# C++ Program For Finding Intersection Point Of Two Linked Lists

There are two singly linked lists in a system. By some programming error, the end node of one of the linked lists got linked to the second list, forming an inverted Y-shaped list. Write a program to get the point where two linked lists merge. Above diagram shows an example with two linked lists having 15 as intersection points.

Method 1(Simply use two loops):
Use 2 nested for loops. The outer loop will be for each node of the 1st list and the inner loop will be for 2nd list. In the inner loop, check if any of the nodes of the 2nd list is the same as the current node of the first linked list. The time complexity of this method will be O(M * N) where m and n are the numbers of nodes in two lists.

Method 2 (Mark Visited Nodes):
This solution requires modifications to basic linked list data structure. Have a visited flag with each node. Traverse the first linked list and keep marking visited nodes. Now traverse the second linked list, If you see a visited node again then there is an intersection point, return the intersecting node. This solution works in O(m+n) but requires additional information with each node. A variation of this solution that doesn’t require modification to the basic data structure can be implemented using a hash. Traverse the first linked list and store the addresses of visited nodes in a hash. Now traverse the second linked list and if you see an address that already exists in the hash then return the intersecting node.

Method 3(Using difference of node counts):

• Get count of the nodes in the first list, let count be c1.
• Get count of the nodes in the second list, let count be c2.
• Get the difference of counts d = abs(c1 – c2)
• Now traverse the bigger list from the first node till d nodes so that from here onwards both the lists have equal no of nodes
• Then we can traverse both the lists in parallel till we come across a common node. (Note that getting a common node is done by comparing the address of the nodes)

Below image is a dry run of the above approach: Below is the implementation of the above approach :

## C++

 `// C++ program to get intersection point of two linked list``#include ``using` `namespace` `std;`` ` `// Link list node ``class` `Node ``{``    ``public``:``    ``int` `data;``    ``Node* next;``};`` ` `// Function to get the counts of ``// node in a linked list ``int` `getCount(Node* head);`` ` `/*  Function to get the intersection ``    ``point of two linked lists head1 ``    ``and head2 where head1 has d more ``    ``nodes than head2 */``int` `_getIntesectionNode(``int` `d, Node* head1,``                        ``Node* head2);`` ` `/* Function to get the intersection point ``   ``of two linked lists head1 and head2 */``int` `getIntesectionNode(Node* head1, ``                       ``Node* head2)``{``    ``// Count the number of nodes in``    ``// both the linked list``    ``int` `c1 = getCount(head1);``    ``int` `c2 = getCount(head2);``    ``int` `d;`` ` `    ``// If first is greater``    ``if` `(c1 > c2) ``    ``{``        ``d = c1 - c2;``        ``return` `_getIntesectionNode(d, head1,``                                   ``head2);``    ``}``    ``else` `    ``{``        ``d = c2 - c1;``        ``return` `_getIntesectionNode(d, head2, ``                                   ``head1);``    ``}``}`` ` `/* Function to get the intersection point ``   ``of two linked lists head1 and head2 ``   ``where head1 has d more nodes than head2 */``int` `_getIntesectionNode(``int` `d, Node* head1, ``                        ``Node* head2)``{``    ``// Stand at the starting of the ``    ``// bigger list``    ``Node* current1 = head1;``    ``Node* current2 = head2;`` ` `    ``// Move the pointer forward``    ``for` `(``int` `i = 0; i < d; i++) ``    ``{``        ``if` `(current1 == NULL) ``        ``{``            ``return` `-1;``        ``}``        ``current1 = current1->next;``    ``}`` ` `    ``// Move both pointers of both list till ``    ``// they intersect with each other``    ``while` `(current1 != NULL && ``           ``current2 != NULL) ``    ``{``        ``if` `(current1 == current2)``            ``return` `current1->data;`` ` `        ``// Move both the pointers forward``        ``current1 = current1->next;``        ``current2 = current2->next;``    ``}`` ` `    ``return` `-1;``}`` ` `/* Takes head pointer of the linked list and ``   ``returns the count of nodes in the list */``int` `getCount(Node* head)``{``    ``Node* current = head;`` ` `    ``// Counter to store count of nodes``    ``int` `count = 0;`` ` `    ``// Iterate till NULL``    ``while` `(current != NULL) ``    ``{``        ``// Increase the counter``        ``count++;`` ` `        ``// Move the Node ahead``        ``current = current->next;``    ``}`` ` `    ``return` `count;``}`` ` `// Driver Code``int` `main()``{``    ``/* Create two linked lists     ``        ``1st 3->6->9->15->30 ``        ``2nd 10->15->30     ``        ``15 is the intersection point */``    ``Node* newNode;`` ` `    ``// Addition of new nodes``    ``Node* head1 = ``new` `Node();``    ``head1->data = 10;`` ` `    ``Node* head2 = ``new` `Node();``    ``head2->data = 3;`` ` `    ``newNode = ``new` `Node();``    ``newNode->data = 6;``    ``head2->next = newNode;`` ` `    ``newNode = ``new` `Node();``    ``newNode->data = 9;``    ``head2->next->next = newNode;`` ` `    ``newNode = ``new` `Node();``    ``newNode->data = 15;``    ``head1->next = newNode;``    ``head2->next->next->next = newNode;`` ` `    ``newNode = ``new` `Node();``    ``newNode->data = 30;``    ``head1->next->next = newNode;`` ` `    ``head1->next->next->next = NULL;`` ` `    ``cout << ``"The node of intersection is "` `<< ``             ``getIntesectionNode(head1, head2);``}``// This code is contributed by rathbhupendra`

Output:

`The node of intersection is 15`

Time Complexity: O(m+n)
Auxiliary Space: O(1)

Method 4(Make circle in first list):
Thanks to Saravanan Man for providing below solution.
1. Traverse the first linked list(count the elements) and make a circular linked list. (Remember the last node so that we can break the circle later on).
2. Now view the problem as finding the loop in the second linked list. So the problem is solved.
3. Since we already know the length of the loop(size of the first linked list) we can traverse those many numbers of nodes in the second list, and then start another pointer from the beginning of the second list. we have to traverse until they are equal, and that is the required intersection point.
4. remove the circle from the linked list.

Time Complexity: O(m+n)
Auxiliary Space: O(1)

Method 5 (Reverse the first list and make equations):
Thanks to Saravanan Mani for providing this method.

```1) Let X be the length of the first linked list until intersection point.
Let Y be the length of the second linked list until the intersection point.
Let Z be the length of the linked list from the intersection point to End of
the linked list including the intersection node.
We Have
X + Z = C1;
Y + Z = C2;
3) Traverse Second linked list. Let C3 be the length of second list - 1.
Now we have
X + Y = C3
We have 3 linear equations. By solving them, we get
X = (C1 + C3 – C2)/2;
Y = (C2 + C3 – C1)/2;
Z = (C1 + C2 – C3)/2;
WE GOT THE INTERSECTION POINT.

Time complexity: O(m+n)
Auxiliary Space: O(1)

Method 6 (Traverse both lists and compare addresses of last nodes): This method is only to detect if there is an intersection point or not. (Thanks to NeoTheSaviour for suggesting this)

```1) Traverse the list 1, store the last node address
2) Traverse the list 2, store the last node address.
3) If nodes stored in 1 and 2 are same then they are intersecting.```

The time complexity of this method is O(m+n) and used Auxiliary space is O(1)

Method 7( 2-pointer technique ):

Using Two pointers :

• Initialize two pointers ptr1 and ptr2  at the head1 and  head2.
• Traverse through the lists,one node at a time.
• When ptr1 reaches the end of a list, then redirect it to the head2.
• similarly when ptr2 reaches the end of a list, redirect it the head1.
• Once both of them go through reassigning, they will be equidistant from
the collision point
• If at any node ptr1 meets ptr2, then it is the intersection node.
• After second iteration if there is no intersection node it returns NULL.

## C++

 `// C++ program to print intersection of lists``#include ``using` `namespace` `std;`` ` `// Link list node ``class` `Node ``{``    ``public``:``    ``int` `data;``    ``Node* next;``};`` ` `// A utility function to return  ``// intersection node``Node* intersectPoint(Node* head1, ``                     ``Node* head2)``{``    ``// Maintaining two pointers ptr1 ``    ``// and ptr2 at the head of A and B,``    ``Node* ptr1 = head1;``    ``Node* ptr2 = head2;`` ` `    ``// If any one of head is NULL i.e``    ``// no Intersection Point``    ``if` `(ptr1 == NULL || ptr2 == NULL) ``    ``{``        ``return` `NULL;``    ``}`` ` `    ``// Traverse through the lists until ``    ``// they reach Intersection node``    ``while` `(ptr1 != ptr2) {`` ` `        ``ptr1 = ptr1->next;``        ``ptr2 = ptr2->next;`` ` `        ``// If at any node ptr1 meets ptr2, ``        ``// then it is intersection node.``        ``// Return intersection node.`` ` `        ``if` `(ptr1 == ptr2) ``        ``{``            ``return` `ptr1;``        ``}`` ` `        ``/* Once both of them go through ``           ``reassigning, they will be ``           ``equidistant from the collision``           ``point.*/`` ` `        ``// When ptr1 reaches the end of a list,``        ``// then reassign it to the head2.``        ``if` `(ptr1 == NULL) ``        ``{``            ``ptr1 = head2;``        ``}``        ``// When ptr2 reaches the end of a list, ``        ``// then redirect it to the head1.``        ``if` `(ptr2 == NULL) ``        ``{``            ``ptr2 = head1;``        ``}``    ``}`` ` `    ``return` `ptr1;``}`` ` `// Function to print intersection nodes``// in  a given linked list``void` `print(Node* node)``{``    ``if` `(node == NULL)``        ``cout << ``"NULL"``;``    ``while` `(node->next != NULL) ``    ``{``        ``cout << node->data << ``"->"``;``        ``node = node->next;``    ``}``    ``cout << node->data;``}`` ` `// Driver code``int` `main()``{``    ``/* Create two linked lists``    ``1st Linked list is 3->6->9->15->30``    ``2nd Linked list is 10->15->30``    ``15 30 are elements in the intersection``    ``list */``    ``Node* newNode;``    ``Node* head1 = ``new` `Node();``    ``head1->data = 10;``    ``Node* head2 = ``new` `Node();``    ``head2->data = 3;``    ``newNode = ``new` `Node();``    ``newNode->data = 6;``    ``head2->next = newNode;``    ``newNode = ``new` `Node();``    ``newNode->data = 9;``    ``head2->next->next = newNode;``    ``newNode = ``new` `Node();``    ``newNode->data = 15;``    ``head1->next = newNode;``    ``head2->next->next->next = newNode;``    ``newNode = ``new` `Node();``    ``newNode->data = 30;``    ``head1->next->next = newNode;``    ``head1->next->next->next = NULL;``    ``Node* intersect_node = NULL;`` ` `    ``// Find the intersection node of``    ``// two linked lists``    ``intersect_node = intersectPoint(head1, ``                                    ``head2);`` ` `    ``cout << ``"INTERSEPOINT LIST :"``;``    ``print(intersect_node);``    ``return` `0;``    ``// This code is contributed by bolliranadheer``}`

Output:

`INTERSEPOINT LIST :15->30`

Time complexity : O( m + n )
Auxiliary Space:  O(1)

Please refer complete article on Write a function to get the intersection point of two Linked Lists for more details!