049.md

160. Intersection of Two Linked Lists

Difficulty: 🟢 Easy

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.

Examples:

Example 1:

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.

Example 2:

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.

Example 3:

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.
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 * 104
• 1 <= Node.val <= 105
• 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?

Solutions

O(m+n) solution

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> Optional[ListNode]:
        pointer_one = headA
        pointer_two = headB

        while pointer_one != pointer_two:
            pointer_one = pointer_one.next if pointer_one else headB
            pointer_two = pointer_two.next if pointer_two else headA

        return pointer_one

The given solution solves the problem using a two-pointer approach. We maintain two pointers, pointer_one and pointer_two, initialized to headA and headB, respectively. We iterate over the linked lists by moving both pointers forward one step at a time. When either pointer reaches the end of its respective linked list, we reset it to the head of the other linked list. We continue this process until both pointers meet at the intersection point or reach the end of the linked lists.

The algorithm follows these steps:

1. Initialize pointer_one to headA and pointer_two to headB.
2. Iterate while pointer_one is not equal to pointer_two:
   • If pointer_one is not None, move it one step forward (pointer_one = pointer_one.next).
   • If pointer_one is None, reset it to headB.
   • If pointer_two is not None, move it one step forward (pointer_two = pointer_two.next).
   • If pointer_two is None, reset it to headA.
3. Return pointer_one, which is the node at which the two linked lists intersect.

Complexity Analysis

The time complexity of the algorithm is O(m + n) because in the worst case, we need to traverse both linked lists to find the intersection point.

The space complexity of the algorithm is O(1) because it uses a constant amount of additional space to store the two pointers.

Summary

The given solution finds the node at which two singly linked lists headA and headB intersect. It uses a two-pointer approach to iterate over the linked lists, resetting the pointers to the heads of the other linked lists when they reach the end. The algorithm has a time complexity of O(m + n) and a space complexity of O(1).

NB: If you want to get community points please suggest solutions in other languages as merge requests.