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.

求一个单链表的中间节点，这题做过类似题目的人很容易就想出使用快慢指针来解决：

struct ListNode {
    int val;
    ListNode *next;
    ListNode() : val(0), next(nullptr) {}
    ListNode(int x) : val(x), next(nullptr) {}
    ListNode(int x, ListNode *next) : val(x), next(next) {}
};

class Solution {
  public:
    ListNode *middleNode(ListNode *head) {
        if (nullptr == head) {
            return nullptr;
        }
        ListNode *slow = head;
        ListNode *quick = head;

        while (slow != nullptr && quick != nullptr) {
            if (quick->next == nullptr || quick->next->next == nullptr) {
                if (quick->next == nullptr) {
                    return slow;
                }
                return slow->next;
            }
            quick = quick->next->next;
            slow = slow->next;
        }
        return slow;
    }
};