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leetcode 160. Intersection of Two Linked Lists (Python)

07 Jun 2020 Leetcode Linked-List

160. Intersection of Two Linked Lists (Python)

Linked-List.

Description

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

160 sample

Sample I/O

Example 1

example1

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

example1

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

example1

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.

Note

  • 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.

Methodology

Once two linked list are intersected, the rest parts of both will be same to end. So we append l2 to l1 and append l1 to l2, then the intersected parts l1l2 and l2l1 has same length and there will be a intersected node at same posiion of both linked list to make the rest part intersected.

Code

def getIntersectionNode(self, headA, headB):
        """
        :type head1, head1: ListNode
        :rtype: ListNode
        """
        h1 = headA
        h2 = headB
        
        while h1 != h2:
            if not h1:
                h1 = headB
            else:
                h1 = h1.next
            
            if not h2:
                h2 = headA
            else:
                h2 = h2.next
                
        return h1

BigO

Iterate the both linked lists once, so total time complexity is O(m+n) where m is length of h1 and n is length of h2

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