# Leetcode: Middle of the Linked List

Middle of the Linked List Similar Problems:

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.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

• Solution:
```// Blog link: https://code.dennyzhang.com/middle-of-the-linked-list
// Basic Ideas: fast-slow pointers
// Complexity: Time O(n), Space O(1)
/**
* Definition for singly-linked list.
* type ListNode struct {
*     Val int
*     Next *ListNode
* }
*/
func middleNode(head *ListNode) *ListNode {
if head == nil { return nil }
f, s := head, head
// last node
for f != nil && f.Next != nil {
s = s.Next
f = f.Next.Next
}
return s
}
```

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