## Description

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

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

begin to intersect at node c1.

**Example 1:**

Input:intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3Output:Reference of the node with value = 8Input 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 [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.

**Example 2:**

Input:intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1Output:Reference of the node with value = 2Input 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 [0,9,1,2,4]. 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 = 2Output:nullInput 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.

**Notes:**

- If the two linked lists have no intersection at all, return
`null`

. - The linked lists must retain their original structure after the function returns.
- You may assume there are no cycles anywhere in the entire linked structure.
- Your code should preferably run in O(n) time and use only O(1) memory.

## Explanation

Find from the second list which node exists in the first list

## Python 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) -> ListNode:
visited = set()
while headA != None:
visited.add(headA)
headA = headA.next
while headB != None:
if headB in visited:
return headB
headB = headB.next
return None
```

- Time complexity: O(N).
- Space complexity: O(1).

I found that solution is very popular and helpful : https://www.youtube.com/watch?v=TxjA3ciq95I

/** Java solution ***

public ListNode getIntersectionNode(ListNode headA, ListNode headB) {

if (headA == null || headB == null) return null;

int lenA = getLength(headA), lenB = getLength(headB);

if (lenA > lenB) {

for (int i = 0; i < lenA – lenB; ++i) headA = headA.next;

} else {

for (int i = 0; i < lenB – lenA; ++i) headB = headB.next;

}

while (headA != null && headB != null && headA != headB) {

headA = headA.next;

headB = headB.next;

}

return (headA != null && headB != null) ? headA : null;

}

public int getLength(ListNode head) {

int cnt = 0;

while (head != null) {

++cnt;

head = head.next;

}

return cnt;

}