Intersection of Two Linked Lists

Intersection of Two Linked Lists - 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: The linked lists must retain their original structure after the function returns.

Input & Output

Example 1 — Lists Intersect

$ Input: listA = [4,1,8,4,5], listB = [5,6,1,8,4,5]

› Output: [8,4,5]

💡 Note: The intersection starts at node with value 8. Both lists share the same tail: 8→4→5

Example 2 — No Intersection

$ Input: listA = [2,6,4], listB = [1,5]

› Output: null

💡 Note: The two linked lists have no common nodes, so return null

Example 3 — Single Node Intersection

$ Input: listA = [3], listB = [2,3]

› Output: [3]

💡 Note: Lists intersect at the single node with value 3

Constraints

The number of nodes in listA is m
The number of nodes in listB is n
1 ≤ m, n ≤ 3 × 10⁴
1 ≤ Node.val ≤ 10⁵

Visualization

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Asked in

a Amazon 65 M Microsoft 45 G Google 38 f Facebook 32

The key insight is that if two linked lists intersect, walking both complete paths gives the same total distance. The optimal two pointers approach has each pointer switch to the other list when reaching the end, guaranteeing they meet at the intersection. Time: O(m+n), Space: O(1)

Common Approaches

✓ Binary Search

⏱️ Time: N/A Space: N/A

Brute Force

⏱️ Time: O(m×n) Space: O(1)

Check every node in the first list against every node in the second list. When we find two nodes that are the same reference (not just same value), that's our intersection point.

Hash Set Lookup

⏱️ Time: O(m+n) Space: O(m)

First pass: traverse list A and store all node references in a hash set. Second pass: traverse list B and check if each node exists in the hash set. First match found is the intersection.

Two Pointers (Optimal)

⏱️ Time: O(m+n) Space: O(1)

Two pointers traverse the lists. When pointer A reaches the end, it starts from headB. When pointer B reaches the end, it starts from headA. They will meet at intersection or both become null if no intersection exists.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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