Solution 1 - Slow & Fast Pointers

Fast moves 2 steps, slow moves 1 step. When fast ends, slow is at the middle (second middle for even length).

TimeComplexity: O(n)

SpaceComplexity: O(1)

""" 0876.1 - Middle of the Linked List - Solution 1 - Slow & Fast Pointers """

from typing import Optional

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

class Solution:
    def middleNode(self, head: Optional[ListNode]) -> Optional[ListNode]:
        slow = fast = head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
        return slow

package main

type ListNode struct {
	Val  int
	Next *ListNode
}

func middleNode(head *ListNode) *ListNode {
	slow, fast := head, head
	for fast != nil && fast.Next != nil {
		slow = slow.Next
		fast = fast.Next.Next
	}
	return slow
}

func main() {}

0876 - Middle of the Linked List

Solution 1 - Slow & Fast Pointers