876. Middle of the Linked List

题目描述：

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.

解题思路：

寻找链表的中间节点，首先看看中间节点有什么特点，中间节点的到头节点的节点数等于尾节点到中间节点的节点数，或者比尾节点到中间节点的节点数多一。

此时可以申明两个节点指向头结点，一个快节点每步走两个节点，一个慢节点每步走一个节点，当快捷点为NULL或快捷点的next节点为NULL时，慢节点指向中间节点。

代码：

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode* middleNode(ListNode* head) {
12         if (head == NULL)
13             return head;
14         ListNode* fast = head;
15         ListNode* slow = head;
16         while (fast != NULL && fast->next != NULL) {
17             fast = fast->next;
18             fast = fast->next;
19             slow = slow->next;
20         }
21         return slow;
22     }
23 };