6.Point of insertion between two Linked List

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:

It is guaranteed 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.

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.

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.

Solution:

class Solution
{
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB)
    {

        if (headA == NULL || headB == NULL)
        {
            return NULL;
        }
        else if (headA == headB)
        {
            return headA;
        }

        ListNode *p = headA;
        ListNode *q = headB;
        ListNode *temp1 = headA;
        ListNode *temp2 = headB;
        ListNode *res = NULL;
        int s = 0, c = 0;

        while (temp1 != NULL)
        {
            s = s + 1;
            temp1 = temp1->next;
        }

        while (temp2 != NULL)
        {
            c = c + 1;
            temp2 = temp2->next;
        }

        temp1 = NULL;
        temp2 = NULL;
        int dif = abs(s - c);
        if (dif == 0)
        {
            while (p != NULL && q != NULL)
            {
                temp1 = p;
                temp2 = q;
                if (temp1 == temp2)
                {
                    res = temp1;
                    break;
                }
                p = p->next;
                q = q->next;
            }
        }

        else
        {
            int i = 0;
            if (s > c)
            {
                while (i < dif)
                {
                    p = p->next;
                    i++;
                }

                while (p != NULL && q != NULL)
                {
                    temp1 = p;
                    temp2 = q;
                    if (temp1 == temp2)
                    {
                        res = temp1;
                        break;
                    }
                    p = p->next;
                    q = q->next;
                }
            }
            else
            {
                while (i < dif)
                {
                    q = q->next;
                    i++;
                }

                while (p != NULL && q != NULL)
                {
                    temp1 = p;
                    temp2 = q;
                    if (temp1 == temp2)
                    {
                        res = temp1;
                        break;
                    }
                    p = p->next;
                    q = q->next;
                }
            }
        }

        return res;
    }
};

Solution Explanation:-

• Find the length of the two linked list.

• Find the difference in length of two linked list.

• Iterate the longer linked list till the difference

• Then iterate over the two linked list parallelly and find the position of the intersection.

Time complexity O(N), Space complexity O(1)