middle_ele_of_linked_list.py

#Link to the problem : https://leetcode.com/problems/middle-of-the-linked-list/solution/

# Approach 1 : Fast and slow pointer

# 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
        """
        
        slow = fast = head
        
        while fast and fast.next :
            slow = slow.next
            fast = fast.next.next
            
        return slow


# Approach 2 
#Time Complexity : O(n)
# CODE : 
#Defining Structure
class Node:
    def __init__(self, data):
       self.data = data
       self.next = None
 
 
class LinkedList:
    def __init__(self):
        self.head = None
        self.last_node = None
 
    def append(self, data):
        if self.last_node is None:
            self.head = Node(data)
            self.last_node = self.head
        else:
            self.last_node.next = Node(data)
            self.last_node = self.last_node.next
 
 
#Function to find the middle element
def FindMiddle(list):
    current = list.head
    length = 0
    #loop to find total length
    while current:
        current = current.next
        length = length + 1
 
    current = list.head
    for i in range((length - 1)//2):
        current = current.next
 
    if current:
    #if even elements then select latter
        if length % 2 == 0:
            print(current.next.data)
        else:
            print(current.data)
    else:
        print('No data found')
 
 
MyList = LinkedList()
 
MyList.append(10)
MyList.append(20)
MyList.append(30)
MyList.append(40)
MyList.append(50)
MyList.append(60)
 
FindMiddle(MyList)