getIntersectionNodeWithHashSet<T> function

LinkedListNode<T>? getIntersectionNodeWithHashSet<T>(LinkedListNode<T>? headA, LinkedListNode<T>? headB)

Finds the intersection point using hash set approach

headA - The head of the first linked list headB - The head of the second linked list Returns the intersection node, or null if no intersection exists

Time Complexity: O(n + m) where n and m are the lengths of the lists Space Complexity: O(n) due to hash set usage

Implementation

LinkedListNode<T>? getIntersectionNodeWithHashSet<T>(
  LinkedListNode<T>? headA,
  LinkedListNode<T>? headB,
) {
  if (headA == null || headB == null) return null;

  Set<LinkedListNode<T>> visited = {};
  LinkedListNode<T>? current = headA;

  // Add all nodes from first list to set
  while (current != null) {
    visited.add(current);
    current = current.next;
  }

  // Check second list for intersection
  current = headB;
  while (current != null) {
    if (visited.contains(current)) {
      return current;
    }
    current = current.next;
  }

  return null;
}