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.

題目分析:
本題要求中間節(jié)點，那么最直接的思路就是先從頭遍歷該鏈表，統(tǒng)計出鏈表長度n，那么中間的節(jié)點就應該是第(n+1)/2個，那么在從列表頭往后遍歷到第(n+1)/2個就是所求的中間節(jié)點。
雖然這種方法的時間復雜度為O(n)，但是我們需要兩次遍歷該鏈表。

一種比較方便的做法是利用快慢指針，這也是在鏈表的問題中比較常見的一種策略。在本題中慢指針每次前進一步，快指針每次前進兩步，這樣在快指針到達鏈表末尾的時候，慢指針就會指向中間節(jié)點。這里注意快慢指針在一開始時都應該初始化為頭結點。這種做法相比于第一種直觀做法，雖然時間復雜度還是O(n)，但算法只需要一次列表遍歷就可以了。

對應的Golang代碼如下

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func middleNode(head *ListNode) *ListNode {
    if head == nil || head.Next == nil {
        return head
    }
    var slow *ListNode = head
    var fast *ListNode = head
    for fast != nil && fast.Next != nil {
        fast = fast.Next.Next
        slow = slow.Next
    }
    return slow
}

對應的C#代碼如下

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode MiddleNode(ListNode head) {
        if (head == null || head.next == null) return head;
            var slow = head;
            var fast = head;
            while (fast != null && fast.next != null)
            {
                slow = slow.next;
                fast = fast.next.next;
            }
            return slow;
    }
}

對應的C++代碼如下

/**
 * 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) {
        if (head == NULL || head->next == NULL) return head;
        ListNode *slow = head, *fast = head;
        while (fast != NULL && fast->next != NULL) {
            slow = slow->next;
            fast = fast->next;
            if (fast != NULL) fast = fast->next;
        }
        return slow;
    }
};

對應的java代碼如下

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode middleNode(ListNode head) {
        ListNode slow = head, fast = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next;
            if (fast.next != null) fast = fast.next;
        }
        return slow;
    }
}