:orphan:

Intersection Of Two Linked Lists
================================

.. highlight:: none

Problem
-------
https://leetcode.com/problems/intersection-of-two-linked-lists/

Given the heads of two singly linked-lists ``headA`` and ``headB``,
return *the node at which the two lists intersect*. If the two linked
lists have no intersection at all, return ``null``.

For example, the following two linked lists begin to intersect at node
``c1``:

|image1|

The test cases are generated such that there are no cycles anywhere in
the entire linked structure.

**Note** that the linked lists must **retain their original structure**
after the function returns.

**Custom Judge:**

The inputs to the **judge** are given as follows (your program is
**not** given these inputs):

- ``intersectVal`` - The value of the node where the intersection
  occurs. This is ``0`` if there is no intersected node.
- ``listA`` - The first linked list.
- ``listB`` - The second linked list.
- ``skipA`` - The number of nodes to skip ahead in ``listA`` (starting
  from the head) to get to the intersected node.
- ``skipB`` - The number of nodes to skip ahead in ``listB`` (starting
  from the head) to get to the intersected node.

The judge will then create the linked structure based on these inputs
and pass the two heads, ``headA`` and ``headB`` to your program. If you
correctly return the intersected node, then your solution will be
**accepted**.

 

**Example 1:**

|image2|

::


   Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,6,1,8,4,5], skipA = 2, skipB = 3
   Output: Intersected at '8'
   Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect).
   From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,6,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.
   - Note that the intersected node's value is not 1 because the nodes with value 1 in A and B (2nd node in A and 3rd node in B) are different node references. In other words, they point to two different locations in memory, while the nodes with value 8 in A and B (3rd node in A and 4th node in B) point to the same location in memory.

**Example 2:**

|image3|

::


   Input: intersectVal = 2, listA = [1,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
   Output: Intersected at '2'
   Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect).
   From the head of A, it reads as [1,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.

**Example 3:**

|image4|

::


   Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
   Output: No intersection
   Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.
   Explanation: The two lists do not intersect, so return null.

 

**Constraints:**

- The number of nodes of ``listA`` is in the ``m``.
- The number of nodes of ``listB`` is in the ``n``.
- ``1 <= m, n <= 3 * 10``\ :sup:```4```
- ``1 <= Node.val <= 10``\ :sup:```5```
- ``0 <= skipA <= m``
- ``0 <= skipB <= n``
- ``intersectVal`` is ``0`` if ``listA`` and ``listB`` do not intersect.
- ``intersectVal == listA[skipA] == listB[skipB]`` if ``listA`` and
  ``listB`` intersect.

 

**Follow up:** Could you write a solution that runs in ``O(m + n)`` time
and use only ``O(1)`` memory?

.. |image1| image:: https://assets.leetcode.com/uploads/2021/03/05/160_statement.png
.. |image2| image:: https://assets.leetcode.com/uploads/2021/03/05/160_example_1_1.png
.. |image3| image:: https://assets.leetcode.com/uploads/2021/03/05/160_example_2.png
.. |image4| image:: https://assets.leetcode.com/uploads/2021/03/05/160_example_3.png

.. highlight:: python

Pattern
-------
Hash Table, Linked List, Two Pointers

Solution
--------
Use a set to store the nodes in the first linked list with ``headA``. Then,
check if any node in the second linked list with ``headB`` is in the set.

Code
----

.. literalinclude:: ../problems/easy/intersection-of-two-linked-lists/intersection_of_two_linked_lists__approach_1.py
    :language: python
    :lines: 17-

Test
----
>>> from intersection_of_two_linked_lists__approach_1 import ListNode, getIntersectionNode
>>> a = ListNode.from_list([4, 1, 8, 4, 5])
>>> b = ListNode(5, ListNode(0, ListNode(1, a.next.next)))
>>> getIntersectionNode(a, b).val
8
>>> a = ListNode.from_list([0, 9, 1, 2, 4])
>>> b = ListNode(3, a.next.next.next)
>>> getIntersectionNode(a, b).val
2
>>> a = ListNode.from_list([2, 6, 4])
>>> b = ListNode.from_list([1, 5])
>>> getIntersectionNode(a, b) is None
True

Complexity
----------
| :math:`m` is the length of the first linked list and :math:`n` is the length of the second linked list
| Time: :math:`O(m + n)` — single pass through both lists
| Space: :math:`O(m)` — set for first list

.. autoclass:: intersection_of_two_linked_lists__approach_1.ListNode
   :members:
   :show-inheritance:
   :undoc-members:

.. autofunction:: intersection_of_two_linked_lists__approach_1.getIntersectionNode