 # 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):
"""
: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
"""