[leetcode-876-Middle of the Linked List]

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.

思路：

快慢指针即可。

struct ListNode
{
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};
ListNode* middleNode(ListNode* head)
{
    ListNode* first = head;
    ListNode* second = head;
    while(second != NULL && second->next != NULL)
    {
        first = first->next;
        second = second->next->next;
    }
    return first;     
}