Find Middle Node of Linked List

function findMiddleNode(head){
  let slow = head, fast = head;
  while(fast && fast.next){ slow = slow.next; fast = fast.next.next; }
  return slow;
}
                  

Explanation: Move fast pointer twice as fast as slow pointer. When fast reaches end, slow is at middle.

⏱ O(n) | 💾 O(1)

class Solution {
  public ListNode findMiddleNode(ListNode head){
    ListNode slow = head, fast = head;
    while(fast != null && fast.next != null){ slow = slow.next; fast = fast.next.next; }
    return slow;
  }
}
                  

Explanation: Floyd’s two-pointer approach to find the middle node in one pass.

⏱ O(n) | 💾 O(1)

def find_middle_node(head):
    slow = fast = head
    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next
    return slow
                  

Explanation: Use slow/fast pointers. Slow reaches middle when fast reaches end.

⏱ O(n) | 💾 O(1)