## # 160. 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 = 3
Output: Reference of the node with value = 8
Input 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 = 1
Output: Reference of the node with value = 2
Input 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 = 2
Output: null
Input 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.

### # Solution

Approach 1: Hash table.

Approach 2: Two pointers: lenA + lenB == lenB + lenA.

### # Code (Python)

Approach 2:

``````    def getIntersectionNode(self, headA, headB):
return None
while True:
if node1 == node2:
return node1
if not node1.next and not node2.next:
return None
if node1.next:
node1 = node1.next
else:
if node2.next:
node2 = node2.next
else:
``````

### # Code (C++)

Approach 1:

``````/**
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
unordered_set<ListNode*> visited;
for (ListNode *p = headA; p != NULL; p = p->next)
{
visited.insert(p);
}
for (ListNode *p = headB; p != NULL; p = p->next)
{
if (visited.find(p) != visited.end())
return p;
}
return NULL;
}
};
``````

Approach 2:

``````class Solution {
public:
while (pA || pB)
{
if (pA == pB)
return pA;
if (pA)
pA = pA->next;
else