# 876. Middle of the Linked List

# Problem

Given the `head`

of a singly linked list, return *the middle node of the linked list*.

If there are two middle nodes, return **the second middle** node.

**Example 1:**

**Input:** head = [1,2,3,4,5] **Output:** [3,4,5] **Explanation:** The middle node of the list is node 3.

**Example 2:**

**Input:** head = [1,2,3,4,5,6] **Output:** [4,5,6] **Explanation:** Since the list has two middle nodes with values 3 and 4, we return the second one.

**Constraints:**

- The number of nodes in the list is in the range
`[1, 100]`

. `1 <= Node.val <= 100`

# Solution

```
# 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
"""
"""
We can do this with 2 loops, can we do it with just 1 loop?
I am thinking 2 pointers
First pointer is "fast", it travels at N speed
Second pointer travels at N/2 speed
So it only increments when pointer1 is > n/2 away
Because by definition the "middle" is half of the max length
therefore the maxium the second pointer can travel is n / 2
Alternatively we can view this a different way. The maximum speed pointer1 can travel is N
the maximum speed pointer2 can travel is half of N, so n / 2
Therefore we use a fast and slow approach
"""
fast, slow = head, head
while fast and fast.next:
# slow will happen to be in the middle when fast can no longer run
fast = fast.next.next
slow = slow.next
return slow
```

This is similar to the fast and slow problem: