Intersection of Two Linked Lists
Tags: leetcode, dsa, theabbie
Published: 2022-06-08 18:30:37

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`:

![](https://assets.leetcode.com/uploads/2021/03/05/160_statement.png)

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:**

![](https://assets.leetcode.com/uploads/2021/03/05/160_example_1_1.png)

**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.

**Example 2:**

![](https://assets.leetcode.com/uploads/2021/03/05/160_example_2.png)

**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:**

![](https://assets.leetcode.com/uploads/2021/03/05/160_example_3.png)

**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 * 104`
*   `1 <= Node.val <= 105`
*   `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?**SOLUTION:**

    # Definition for singly-linked list.
    # class ListNode:
    #     def __init__(self, x):
    #         self.val = x
    #         self.next = None
    
    class Solution:
        def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> Optional[ListNode]:
            if headA and headB:
                a = headA
                b = headB
                while a != b:
                    a = a.next if a else headB
                    b = b.next if b else headA
                return a
            else:
                return None