# 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`:![](https://assets.leetcode.com/uploads/2021/03/05/160_statement.png)

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:**

![](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.
```

## Solution: 

```cpp
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:-**&#x20;

* Find the length of the two linked list.
* Find the difference in length of two linked list.
* &#x20;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)**


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