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Suppose we have two linked lists A and B, there are few elements in these linked lists. We have to return the reference of the intersection points. The inputs are intersectionVal = 8, A = [4,1,8,4,5], B = [5,0,1,8,4,5], skipA = 2 and skipB = 3, these are used to skip 2 elements from A and skip 3 elements from B.

To solve this, we will follow these steps −

- Define a map called d
- while headA is not null
- d[headA] := 1
- headA := next of headA

- while headB is not null
- if headB in d
- return headB

- headB := next of headB

- if headB in d
- Return null

Let us see the following implementation to get better understanding −

class ListNode: def __init__(self, data, next = None): self.data = data self.next = next class Solution(object): def getIntersectionNode(self, headA, headB): """ :type head1, head1: ListNode :rtype: ListNode """ dict = {} while headA: dict[headA]=1 headA = headA.next while headB: if headB in dict: return headB headB = headB.next return None headA = ListNode(4) headB = ListNode(5) Intersect = ListNode(8, ListNode(4, ListNode(5))) headA.next = ListNode(1, Intersect) headB.next = ListNode(0, ListNode(1, Intersect)) ob1 = Solution() op = ob1.getIntersectionNode(headA, headB) print("Intersection:",op.data)

headA = ListNode(4) headB = ListNode(5) Intersect = ListNode(8, ListNode(4, ListNode(5))) headA.next = ListNode(1, Intersect) headB.next = ListNode(0, ListNode(1, Intersect))

Intersected at '8'

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