# 0160. Intersection of Two Linked Lists

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

## Description

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`.
* `0 <= 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(n)` time and use only `O(1)` memory?

## ac1: two pass

intuitive yet tedious

```java
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        // edge cases
        if (headA == null || headB == null) return null;

        // intersect? walk 2 list
        int lena = 1, lenb = 1;
        ListNode a = headA, b = headB;
        while (a.next != null) {
            a = a.next;
            lena++;
        }
        while (b.next != null) {
            b = b.next;
            lenb++;
        }
        if (a != b) return null; // last node different, no intersection

        int diff = Math.abs(lena - lenb);
        // let a be the longer one
        if (lena < lenb) { // swap
            a = headB;
            b = headA;
        } else {
            a = headA;
            b = headB;
        }

        // cut diff in longer one
        for (int i = 0; i < diff; i++) {
            a = a.next;
        }

        // same length, find intersection
        while (a != null && b != null) {
            if (a == b) break;
            a = a.next;
            b = b.next;
        }

        return a;
    }
}
```

## ac2: little trick

```java
public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        // edge cases
        if (headA == null || headB == null) return null;

        // intersect? walk 2 list
        ListNode a = headA, b = headB;

        while (a != b) {  // if a b has not intersection, they will reach null at the same time, thus eaqual
            a = a == null ? headB : a.next;
            b = b == null ? headA : b.next;
        }

        return a;
    }
}
```
