Delete the Middle Node of a List

medium linked list two pointers

Problem

Given the head of a singly linked list, delete the middle node and return the new head. The middle node of a list of length n is at position ⌊n/2⌋ (0-indexed). Use slow and fast pointers — fast moves twice as quickly, so when fast hits the end, slow sits at the middle.

Inputhead = 5 → 8 → 1 → 3 → 9 → null

Output5 → 8 → 3 → 9 → null

Length 5; middle index 2 (value 1) is removed.

node values (comma-separated)

def delete_middle(head):
    if head is None or head.next is None:
        return None
    slow, fast, prev = head, head, None
    while fast and fast.next:
        prev = slow
        slow = slow.next
        fast = fast.next.next
    prev.next = slow.next
    return head

function deleteMiddle(head) {
  if (!head || !head.next) return null;
  let slow = head, fast = head, prev = null;
  while (fast && fast.next) {
    prev = slow;
    slow = slow.next;
    fast = fast.next.next;
  }
  prev.next = slow.next;
  return head;
}

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

ListNode* deleteMiddle(ListNode* head) {
    if (!head || !head->next) return nullptr;
    ListNode *slow = head, *fast = head, *prev = nullptr;
    while (fast && fast->next) {
        prev = slow;
        slow = slow->next;
        fast = fast->next->next;
    }
    prev->next = slow->next;
    return head;
}

Time: O(n) Space: O(1)