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**.

**Example 1:**

Input:intersectVal = 8, listA = [4,1,8,4,5], listB = [5,6,1,8,4,5], skipA = 2, skipB = 3Output: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 (2^{nd}node in A and 3^{rd}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 (3^{rd}node in A and 4^{th}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 = 1Output: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 = 2Output:No intersectionExplanation: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`

. `1 <= m, n <= 3 * 10`

^{4}`1 <= Node.val <= 10`

^{5}`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.

` ````
```public class Solution {
public int sizeLL(ListNode head){
int count = 0 ;
ListNode ptr = head ;
while(ptr!=null){
count++;
ptr = ptr.next;
}
return count ;
}
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
if(headA == null || headB == null){
return null;
}
int size1 = sizeLL(headA);
int size2 = sizeLL(headB);
int diff = size1 - size2 ;
ListNode ptr1 = headA;
ListNode ptr2 = headB;
if(diff>0){ // LL1 is having greater size
while(diff>0){
ptr1 = ptr1.next;
diff--;
}
}else{ // LL2 is having greater size
while(diff<0){
ptr2 = ptr2.next;
diff++;
}
}
while(ptr1!=ptr2){
ptr1 = ptr1.next;
ptr2 = ptr2.next;
}
return ptr1;
}
}

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