## Description

https://leetcode.com/problems/middle-of-the-linked-list/

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:
if head == None:
return head
count = 0
fast = head
while fast != None:
fast = fast.next
count += 1
mid = count // 2 + 1
slow = head
for i in range(1, mid):
slow = slow.next
return slow
```

- Time complexity: O(n).
- Space complexity: O(1).

/**

* 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) {

ListNode end = head;

ListNode mid = head;

while(end!=null && end.next!=null){

end = end.next.next;

mid = mid.next;

}

return mid;

}

public ListNode middleNode1(ListNode head) {

ListNode midNode = head;

int len = getLength(head);

int mid = len/2;

while(mid > 0){

midNode = head.next;

head = head.next;

mid–;

}

return midNode;

}

public int getLength(ListNode head){

int count =0;

while(head != null){

head = head.next;

count++;

}

return count;

}

}

This looks great! To further explain this solution, I write a post with graphical explanation. Hopefully it will helps someone. It’s available at https://medium.com/@edward.zhou/leet-code-876-middle-of-the-linked-list-explained-python3-solution-a9417b418bc7.