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Middle of the Linked List

Easy
Fast & Slow Pointers

Given the head of a singly linked list, return the middle node of the linked list.

If there are two middle nodes, return the second middle node.

This problem elegantly demonstrates the fast & slow pointer technique. When the fast pointer reaches the end, the slow pointer will be at exactly the middle. Think of it like two runners on a track - when the faster one finishes, the slower one is at the halfway point.

Example 1

Input: head = [1,2,3,4,5]
Output: [3,4,5]
Explanation: The middle node of the list is node 3.

Example 2

Input: head = [1,2,3,4,5,6]
Output: [4,5,6]
Explanation: Since the list has two middle nodes with values 3 and 4, we return the second one.

Constraints

  • •The number of nodes in the list is in the range [1, 100]
  • •1 <= Node.val <= 100
Show Hints (4)
Hint 1: If you walk through the list once to count nodes, then walk halfway, that works but requires two passes. Can you do it in one pass?
Hint 2: What if one pointer moves twice as fast as another?
Hint 3: When the fast pointer has traversed the entire list, where would a pointer moving at half the speed be?
Hint 4: For even-length lists, why does stopping when fast.next is null give us the second middle?
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