原文链接：876. 链表的中间结点
给你单链表的头结点 head ，请你找出并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。
示例 1：
输入：head = [1,2,3,4,5]
输出：[3,4,5]
解释：链表只有一个中间结点，值为 3 。
示例 2：
输入：head = [1,2,3,4,5,6]
输出：[4,5,6]
解释：该链表有两个中间结点，值分别为 3 和 4 ，返回第二个结点。
提示：
[1, 100]1 <= Node.val <= 100/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode middleNode(ListNode head) {
if(head == null || head.next == null){
return head;
}
ListNode slow = head;
ListNode fast = head;
while(fast != null && fast.next != null){
fast = fast.next.next;
slow = slow.next;
}
return slow;
}
}
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution(object):
def middleNode(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
if head == None or head.next == None:
return head
slow = head
fast = head
while fast != None and fast.next != None:
fast = fast.next.next
slow = slow.next
return slow
/**
* Definition for singly-linked list.
* 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(head == nullptr || head->next == nullptr){
return head;
}
ListNode* slow = head;
ListNode* fast = head;
while(fast != nullptr && fast->next != nullptr){
fast = fast->next->next;
slow = slow->next;
}
return slow;
}
};
/**
* 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
}
slow := head
fast := head
for fast != nil && fast.Next != nil {
fast = fast.Next.Next
slow = slow.Next
}
return slow
}