Python program to find middle of a linked list using one traversal

Given a singly linked list, find the middle of the linked list. Given a singly linked list, find the middle of the linked list. For example, if the given linked list is 1->2->3->4->5 then the output should be 3.

Method 1: Traverse the whole linked list and count the no. of nodes. Now traverse the list again till count/2 and return the node at count/2. 

Python3

# Python program for the above approach
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None


class NodeOperation:
    # Function to add a new node
    def pushNode(self, head_ref, data_val):

        # Allocate node and put in the data
        new_node = Node(data_val)

        # Link the old list of the new node
        new_node.next = head_ref

        # move the head to point to the new node
        head_ref = new_node
        return head_ref

    # A utility function to print a given linked list
    def printNode(self, head):
        while (head != None):
            print('%d->' % head.data, end="")
            head = head.next
        print("NULL")

    ''' Utility Function to find length of linked list '''

    def getLen(self, head):
        temp = head
        len = 0

        while (temp != None):
            len += 1
            temp = temp.next

        return len

    def printMiddle(self, head):
        if head != None:
            # find length
            len = self.getLen(head)
            temp = head

            # traverse till we reached half of length
            midIdx = len // 2
            while midIdx != 0:
                temp = temp.next
                midIdx -= 1

            # temp will be storing middle element
            print('The middle element is: ', temp.data)


# Driver Code
head = None
temp = NodeOperation()
head = temp.pushNode(head, 5)
head = temp.pushNode(head, 4)
head = temp.pushNode(head, 3)
head = temp.pushNode(head, 2)
head = temp.pushNode(head, 1)
temp.printNode(head)
temp.printMiddle(head)

# This code is contributed by Yash Agarwal

Method 2: Traverse linked list using two pointers. Move one pointer by one and another pointer by two. When the fast pointer reaches the end slow pointer will reach middle of the linked list. 

Implementation:

Python3

# Python 3 program to find the middle of a
# given linked list

# Node class
class Node:

    # Function to initialise the node object
    def __init__(self, data):
        self.data = data
        self.next = None

class LinkedList:

    def __init__(self):
        self.head = None

    def push(self, new_data):
        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node

    # Function to get the middle of
    # the linked list
    def printMiddle(self):
        slow_ptr = self.head
        fast_ptr = self.head

        if self.head is not None:
            while (fast_ptr is not None and fast_ptr.next is not None):
                fast_ptr = fast_ptr.next.next
                slow_ptr = slow_ptr.next
            print("The middle element is: ", slow_ptr.data)

# Driver code
list1 = LinkedList()
list1.push(5)
list1.push(4)
list1.push(2)
list1.push(3)
list1.push(1)
list1.printMiddle()

Output

The middle element is:  2

Method 3: Initialized the temp variable as head Initialized count to Zero Take loop till head will become Null(i.e end of the list) and increment the temp node when count is odd only, in this way temp will traverse till mid element and head will traverse all linked list. Print the data of temp. 

Implementation:

Python3

# Python 3 program to find the middle of a
# given linked list

class Node:
    def __init__(self, value):
        self.data = value
        self.next = None

class LinkedList:

    def __init__(self):
        self.head = None

    # create Node and make linked list
    def push(self, new_data):
        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node

    def printMiddle(self):
        temp = self.head
        count = 0

        while self.head:

            # only update when count is odd
            if (count & 1):
                temp = temp.next
            self.head = self.head.next

            # increment count in each iteration
            count += 1

        print(temp.data)

# Driver code
llist = LinkedList()
llist.push(1)
llist.push(20)
llist.push(100)
llist.push(15)
llist.push(35)
llist.printMiddle()
# code has been contributed by - Yogesh Joshi

Output

100

Complexity Analysis:

• Time complexity: O(N) where N is the size of the given linked list
• Auxiliary Space: O(1) because it is using constant space