876. Middle of the Linked List

问题
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.

快慢指针，两个指针不同速度遍历链表。当快指针达到链表尾部时候，慢指针正好在中间。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode middleNode(ListNode head) {
        
        ListNode slowNode = head;
        ListNode fastNode = head;
        
        while( fastNode != null ){
            fastNode = fastNode.next;
            if (fastNode == null){
                return slowNode;
            }
            slowNode = slowNode.next;
            fastNode = fastNode.next;
        }
        return slowNode;
    }
}