# Leetcode Intersection-of-Two-Linked-Lists

2022-09-08 16:13

> ##### Algorithms:
>
> #algorithm #two_pointers
>
> ##### Data structures:
>
> #DS #linked_list #set
>
> ##### Difficulty:
>
> #coding_problem #difficulty-easy
>
> ##### Additional tags:
>
> #leetcode
>
> ##### Revisions:
>
> N/A

##### Links:

- [Link to problem](https://leetcode.com/problems/intersection-of-two-linked-lists/)

---

### Problem

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**.

#### Examples

**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.

- 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:**

![](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.

### Thoughts

> [!summary]
> This can be solved using #two_pointers or #set

#### Set

Record the address of visited node iterating over `listA`,
and verify if the address has been visited when iterating
over `listB`

#### Two pointers

#TODO Use two pointers method to solve

### Solution

#### Set

```cpp
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
  ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
    // O(M + N), using set.
    unordered_set<ListNode *> used;

    ListNode *ptrA = headA;

    while (ptrA) {
      used.insert(ptrA);
      ptrA = ptrA->next;
    }

    ListNode *ptrB = headB;

    while (ptrB) {
      if (used.find(ptrB) != used.end()) {
        // intersect
        return ptrB;
      }

      ptrB = ptrB->next;
    }

    return nullptr;
  }
};
```