# LeetCode 876. Middle of the Linked List

## Description

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`.

## Explanation

fast pointer to find the middle number. slow pointer to get to the middle node.

## Python Solution

``````# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
def middleNode(self, head: ListNode) -> ListNode:

count = 0

while fast != None:
fast = fast.next
count += 1

mid = count // 2 + 1

for i in range(1, mid):
slow = slow.next

return slow``````
• Time complexity: O(n).
• Space complexity: O(1).

## 3 Thoughts to “LeetCode 876. Middle of the Linked List”

1. Vickey says:

/**
* 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(end!=null && end.next!=null){
end = end.next.next;
mid = mid.next;
}
return mid;
}

int mid = len/2;

while(mid > 0){
mid–;
}

return midNode;
}