Formatted question description: https://leetcode.ca/all/876.html

876. Middle of the Linked List (Easy)

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.

Solution 1. Two Pointers

// OJ: https://leetcode.com/problems/middle-of-the-linked-list/
// Time: O(N)
// Space: O(1)
class Solution {
public:
while (fast && fast->next) {
slow = slow->next;
fast = fast->next->next;
}
return slow;
}
};

• /**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
class Solution {
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
}

############

/**
* 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 {
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
}

• // OJ: https://leetcode.com/problems/middle-of-the-linked-list/
// Time: O(N)
// Space: O(1)
class Solution {
public:
while (fast && fast->next) {
slow = slow->next;
fast = fast->next->next;
}
return slow;
}
};

• # Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
def middleNode(self, head: ListNode) -> ListNode:
while fast and fast.next:
slow, fast = slow.next, fast.next.next
return slow

############
3
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
"""
:rtype: ListNode
"""
dummy = ListNode(0)
slow, fast = dummy, dummy
while fast and fast.next:
slow = slow.next
fast = fast.next.next
return slow.next if fast else slow

• /**
* type ListNode struct {
*     Val int
*     Next *ListNode
* }
*/
for fast != nil && fast.Next != nil {
slow, fast = slow.Next, fast.Next.Next
}
return slow
}

• /**
* class ListNode {
*     val: number
*     next: ListNode | null
*     constructor(val?: number, next?: ListNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.next = (next===undefined ? null : next)
*     }
* }
*/

function middleNode(head: ListNode | null): ListNode | null {
while (fast != null && fast.next != null) {
fast = fast.next.next;
slow = slow.next;
}
return slow;
}


• /**
* Definition for a singly-linked list.
* class ListNode {
*     public $val = 0; * public$next = null;
*     function __construct($val = 0,$next = null) {
*         $this->val =$val;
*         $this->next =$next;
*     }
* }
*/
class Solution {
/**
* @param ListNode $head * @return ListNode */ function middleNode($head) {
$count = 0;$tmpHead = $head; while ($tmpHead != null) {
$tmpHead =$tmpHead->next;
$count++; }$len = $count - floor($count / 2);
while ($count !=$len) {
$head =$head->next;
$count--; } return$head;
}
}


• // Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
pub fn middle_node(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {