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.

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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.


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Constraints:


	The number of nodes of listA is in the m.
	The number of nodes of listB is in the n.
	1 &lt;= m, n &lt;= 3 * 104
	1 &lt;= Node.val &lt;= 105
	0 &lt;= skipA &lt;&nbsp;m
	0 &lt;= skipB &lt;&nbsp;n
	intersectVal is 0 if listA and listB do not intersect.
	intersectVal == listA[skipA] == listB[skipB] if listA and listB intersect.


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Follow up: Could you write a solution that runs in O(m + n) time and use only O(1) memory?
class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> Optional[ListNode]:
        m = 0
        n = 0
        temp = headA
        while temp != None:
            m+=1
            temp = temp.next
        temp = headB
        while temp != None:
            n+=1
            temp = temp.next
        diff = 0
        if m>=n : 
            diff = m-n
        else:
            diff = n-m
        p1 = headA
        p2 = headB
        if max(m,n) == m:
            while diff > 0:
                p1 = p1.next
                diff-=1
        else:
            while diff > 0:
                p2 = p2.next
                diff-=1
        while p1 != None and p2!=None:
            if p1 == p2:
                return p1
            p1 = p1.next
            p2 = p2.next
        return None
public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        ListNode tempA = headA, tempB = headB;
        int lenA = 0, lenB = 0;
        while (tempA != null) {
            lenA++;
            tempA = tempA.next;
        }
        while (tempB != null) {
            lenB++;
            tempB = tempB.next;
        }
        tempA = headA; tempB = headB;
        if (lenB > lenA) {
            for (int i = 0; i < lenB - lenA; i++) {
                tempB = tempB.next;
            }
        } else if (lenA > lenB) {
            for (int i = 0; i < lenA - lenB; i++) {
                tempA = tempA.next;
            }
        }
        while (tempA != null && tempA != tempB) {
            tempA = tempA.next;
            tempB = tempB.next;
        }
        return tempA;
    }
}
/**
 * Definition for singly-linked list.
 * struct ListNode {
 * int val;
 * ListNode *next;
 * ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:

    //function to get length of linked list
    int getLength(ListNode *head)
    {
        //initial length 0
        int l = 0;

        while(head != NULL)
        {
            l++;
            head = head->next;
        }
        return l;
    }

    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {

        //ptr1 & ptr2 will move to check nodes are intersecting or not
        //ptr1 will point to the list which have higher length, higher elements (big list)
        //ptr2 will point to small list
        ListNode *ptr1,*ptr2;

        //fetching length of both the list,as we want a node that both ptr points simultaneously
        //so if both list have same length of nodes, they travel with same speed, they intersect
        //if diff legths, it'll be difficult for us to catch
        //so by substracting list with higher length with lower length
        //we get same level of length, from which both ptr start travaeling, they may get intersect
        int length1 = getLength(headA);
        int length2 = getLength(headB);

        int diff = 0;

        //ptr1 points to longer list
        if(length1>length2)
        {
            diff = length1-length2;
            ptr1 = headA;
            ptr2 = headB;

        }
        else
        {
            diff = length2 - length1;
            ptr1 = headB;
            ptr2 = headA;
        }

        //till diff is zero and we reach our desired posn
        while(diff)
        {
            //incrementing ptr1 so that it can be at a place where both list have same length

            ptr1 = ptr1->next;

            if(ptr1 == NULL)
            {
                return NULL;
            }

            diff--;
        }

        //traverse both pointers together, till any of them gets null
        while((ptr1 != NULL) && (ptr2 != NULL))
        {
            //at any point, both point to same node return it
            if(ptr1 == ptr2)
            {
                return ptr1;
            }

            //increment both pointers, till they both be NULL
            ptr1 = ptr1->next;
            ptr2 = ptr2->next;
        }

        //if not found any intersaction, return null
        return NULL;

     }
};
var getIntersectionNode = function(headA, headB) {
   let node1 = headA;
   let node2 = headB;

    const set = new Set()

    while(node1 !== undefined || node2!== undefined){
        if(node1 && set.has(node1)){
            return node1
        }
        if(node2 && set.has(node2)){
            return node2
        }
        if(node1){
            set.add(node1)
            node1 = node1.next;
        }
        else if(node2){
            set.add(node2)
            node2 = node2.next;
        }
        else {return}
    }
 };
Intersection of Two Linked Lists