Day 8: Middle of the Linked List.py

# Day 8: 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.


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

class Solution(object):
    def middleNode(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        faster_pointer = head
        lower_pointer = head

        while faster_pointer and lower_pointer.next:
            if lower_pointer.next.next == None:
                return faster_pointer.next
            faster_pointer = faster_pointer.next
            lower_pointer = lower_pointer.next.next
        return faster_pointer