<gh_stars>1-10
"""
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.

"""


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

class Solution:
    def middleNode(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        """
        Method 1: Traverse with slowPtr and fastPtr

        * Assign slowPtr and fastPtr to head
        * Update the slowPtr by 1 and the fastPtr by 2
        * When the fastPtr reaches the end, the slowPtr
        should have reached the middle of the Linked List.

        Your runtime beats 70.93 % of python3 submissions.
        """
        slowPtr = fastPtr = head
        while fastPtr and fastPtr.next:
            slowPtr = slowPtr.next
            fastPtr = fastPtr.next.next

        return slowPtr
"""
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.

"""


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

class Solution:
    def middleNode(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        """
        Method 1: Traverse with slowPtr and fastPtr

        * Assign slowPtr and fastPtr to head
        * Update the slowPtr by 1 and the fastPtr by 2
        * When the fastPtr reaches the end, the slowPtr
        should have reached the middle of the Linked List.

        Your runtime beats 70.93 % of python3 submissions.
        """
        slowPtr = fastPtr = head
        while fastPtr and fastPtr.next:
            slowPtr = slowPtr.next
            fastPtr = fastPtr.next.next

        return slowPtr
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. # Definition for singly-linked list. # class ListNode: #     def __init__(self, x): #         self.val = x #         self.next = None :type head: ListNode         :rtype: ListNode Method 1: Traverse with slowPtr and fastPtr          * Assign slowPtr and fastPtr to head         * Update the slowPtr by 1 and the fastPtr by 2         * When the fastPtr reaches the end, the slowPtr         should have reached the middle of the Linked List.          Your runtime beats 70.93 % of python3 submissions.