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`: 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:** **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:** **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:** **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?