LeetCode-Middle of the Linked List

Description：
Given a non-empty, singly linked list with head node head, return a middle node of linked list.

If there are two middle nodes, return the second middle node.

Example 1:
Input: [1,2,3,4,5]
Output: Node 3 from this list (Serialization: [3,4,5])
The returned node has value 3. (The judge’s serialization of this node is [3,4,5]).
Note that we returned a ListNode object ans, such that:
ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.

Example 2:
Input: [1,2,3,4,5,6]
Output: Node 4 from this list (Serialization: [4,5,6])
Since the list has two middle nodes with values 3 and 4, we return the second one.

Note:
The number of nodes in the given list will be between 1 and 100.

题意：给定一个链表，要求找出链表的中间节点，如果链表节点数为偶数，则返回中间两个节点的后一个节点；

解法：要找到链表的中间节点，我们就需要知道链表的长度，因此，我们先遍历一遍链表，得出链表的长度后，再遍历一半的链表节点找到链表的中间节点；

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode middleNode(ListNode head) {
        ListNode last = head;
        int len = 0;
        while(last != null){
            len++;
            last = last.next;
        }
        for(int i=0; i < len/2; i++){
            head = head.next;
        }
        return