Intersection of Two Linked Lists

Tested Python solution for LeetCode 160 with 15 pytest cases. Generate a practice environment with lcpy.

LeetCode 160, Easy. Topics: Hash Table, Linked List, Two Pointers. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 15 parametrized pytest cases, and a playground notebook:

lcpy gen -n 160   # by problem number
lcpy gen -s intersection_of_two_linked_lists   # by problem name

​

Problem

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.

​

Examples

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.
- Note that the intersected node's value is not 1 because the nodes with value 1 in A and B (2nd node in A and 3rd 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 (3rd node in A and 4th node in B) point to the same location in memory.

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.

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.

​

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.

Follow up: Could you write a solution that runs in O(m + n) time and use only O(1) memory?

​

Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

from leetcode_py import ListNode


class Solution:
    # Time: O(m + n)
    # Space: O(1)
    def get_intersection_node(
        self, head_a: ListNode[int] | None, head_b: ListNode[int] | None
    ) -> ListNode[int] | None:
        pointer_a: ListNode[int] | None = head_a
        pointer_b: ListNode[int] | None = head_b
        while pointer_a is not pointer_b:
            pointer_a = pointer_a.next if pointer_a is not None else head_b
            pointer_b = pointer_b.next if pointer_b is not None else head_a
        return pointer_a

​

Complexity

Time	Space
O(m + n)	O(1)

​

Tags

NeetCode All.