题目要求 （高频题）
给一个带有头节点head并且非空的单链表。返回链表的中间节点。
如果有两个中间结点，那么返回第二个。
示例
// 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.
解体思路
快慢指针法
和 leetcode 141 判断链表中是否有环 的思路是一样的，通过设置快慢指针，从表头开始遍历，快指针每次走两步，慢指针走一步。
当快指针到达链表尾时。慢指针刚好到达链表中点，返回慢指针即可
/*** Definition for singly-linked list.* struct ListNode {* int val;* ListNode *next;* ListNode(int x) : val(x), next(NULL) {}* };*/
class Solution {
public:ListNode* middleNode(ListNode* head) {
ListNode* slow = head;ListNode* fast = head;while (fast && fast->next) //判断非空条件{
slow = slow->next;fast = fast->next->next;}return slow;}
};