876. Middle of the Linked List

Problem Description

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.

Examples:

Input: head = [1,2,3,4,5]
Output: [3,4,5]
Explanation: The middle node of the list is node 3.

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

Python Solution

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

Implementation Notes:

• Uses fast and slow pointer technique (Floyd's Tortoise and Hare)
• Time complexity: O(n)
• Space complexity: O(1)

Java Solution

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

C++ Solution

class Solution {
public:
    ListNode* middleNode(ListNode* head) {
        ListNode* slow = head;
        ListNode* fast = head;
        
        while (fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
        }
        
        return slow;
    }
};

JavaScript Solution

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var middleNode = function(head) {
    let slow = head;
    let fast = head;
    
    while (fast && fast.next) {
        slow = slow.next;
        fast = fast.next.next;
    }
    
    return slow;
};

C# Solution

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

Implementation Notes:

• Uses two-pointer technique
• Fast pointer moves twice as fast as slow pointer
• When fast reaches end, slow is at middle