题目：

Given the head of a singly linked list, return the middle node of the linked list.

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

给定一个头结点为 head 的非空单链表，返回链表的中间结点。

如果有两个中间结点，则返回第二个中间结点

Example 1:

Input: head = [1,2,3,4,5]
Output: [3,4,5]
Explanation: The middle node of the list is node 3.

示例 1：

输入：[1,2,3,4,5]
输出：此列表中的结点 3 (序列化形式：[3,4,5])
说明：返回的结点值为 3 。 (序列化表述是 [3,4,5])。
注意，我们返回了一个 ListNode 类型的对象 ans，这样：
ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, 以及 ans.next.next.next = NULL.

Example 2:

Input: head = [1,2,3,4,5,6]
Output: [4,5,6]
Explanation: Since the list has two middle nodes with values 3 and 4, we return the second one.

示例 2：

输入：[1,2,3,4,5,6]
输出：此列表中的结点 4 (序列化形式：[4,5,6])
说明：由于该列表有两个中间结点，值分别为 3 和 4，我们返回第二个结点。

Constraints:

• The number of nodes in the list is in the range [1, 100].
• 1 <= Node.val <= 100

提示：

• 列表中的结点数在[1,100]范围内。1 <= Node.val <= 100
• 1 <= Node.val <= 100

解题思路：

方法一：数组

我们可以对链表进行遍历，同时将遍历到的元素依次放入数组 a中。当链表以及数组的长度也为 n，对应的中间节点即为 a[n/2]。

Python代码

class Solution:
    def middleNode(self, head: ListNode) -> ListNode:
        a = [head]
        while a[-1].next:
            a.append(a[-1].next)
        return a[len(a) // 2]

class Solution {
    public ListNode middleNode(ListNode head) {
        ListNode[] a = new ListNode[100];
        int n = 0;
        while (head != null) {
            a[n++] = head;
            head = head.next;
        }
        return a[n / 2];
    }
}

class Solution {
public:
  ListNode* middleNode(ListNode* head) {
	vector<ListNode*> a = { head };
	while (a.back()->next != NULL){
		a.push_back(a.back()->next);
    }
	return a[a.size() / 2];
    }
};

class Solution:
    def middleNode(self, head: ListNode) -> ListNode:
        slow, fast = head, head # slow = fast = head
        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next
        return slow

class Solution {
    public ListNode middleNode(ListNode head) {
        ListNode fast = head;
        ListNode slow = head;
        while(fast != null && fast.next != null){
            fast = fast.next.next;
            slow = slow.next;
        }
        return slow;
    }
}

class Solution {
public:
    ListNode* middleNode(ListNode* head) {
        ListNode* slow = head;
        ListNode* fast = head;
        while (fast != NULL && fast->next != NULL) {
            slow = slow->next;
            fast = fast->next->next;
        }
        return slow;
    }
};