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. 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 Solution Approach 1: Two Passes Find the length of the linked list by traversing the list. Find the middle node by traversing the list again to the middle. Return the middle node # Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next def middle_node(head: ListNode) -> ListNode: length = 0 current = head while current: length += 1 current = current.next middle = length // 2 current = head for _ in range(middle): current = current.next return current Time complexity: O(n) Space complexity: O(1) https://youtu.be/xcsDK3mziCM?embedable=true Approach 2: One Pass Use two pointers, one fast and one slow. Move the fast pointer two nodes at a time and the slow pointer one node at a time. # Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next def middle_node(head: ListNode) -> ListNode: slow = fast = head while fast and fast.next: slow = slow.next fast = fast.next.next return slow Time complexity: O(n) Space complexity: O(1) https://youtu.be/vD0XdyM_Wwk?embedable=true
