Solution: Middle of the Linked List

Let's solve the Middle of the Linked List problem using the Fast and Slow Pointers pattern.


Given the head of a singly linked list, return the middle node of the linked list. If the number of nodes in the linked list is even, there will be two middle nodes, so return the second one.


Let n be the number of nodes in a linked list.

  • 11 \leq n 100\leq 100
  • 11 \leq 100\leq 100
  • head \neq NULL


So far, you have probably brainstormed some approaches and have an idea of how to solve this problem. Let’s explore some of these approaches and figure out which one to follow based on considerations such as time complexity and any implementation constraints.

Naive approach

In the naive approach, we use an external array to store the elements of the linked list, and then we return the element present at the index array.length2\lfloor \frac{array.length}{2} \rfloor as the middle node of the linked list. The time and space complexity of this approach is O(n)O(n), where nn is the number of nodes in the linked list. Let’s see if we can solve this problem with better time and space complexity.

Optimized approach using fast and slow pointers

We can use the fast and slow pointers to solve this problem with constant space complexity. The slow pointer traverses the linked list one step at a time, while the fast pointer takes two steps at a time. This makes the fast pointer reach the end of the linked list in n2\frac{n}{2} iterations, and the slow pointer, by this time, reaches the middle of the linked list.

The following steps are applied when identifying the middle of the linked list:

  1. Create two pointers, slow and fast, initially at the head of the linked list, that is, pointing to the first node.

  2. Traverse the linked list using both pointers, where the slow pointer will move one step forward, and the fast pointer will move two steps forward.

  3. When the fast pointer reaches the last element of the linked list, or becomes equal to NULL, the slow pointer, at that time, will point to the middle node. Return the node that the slow pointer points to.

The slides below help to understand the solution in a better way.

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