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Python Program For Finding The Middle Element Of A Given Linked List

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  • Last Updated : 22 Jun, 2022
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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. 
If there are even nodes, then there would be two middle nodes, we need to print the second middle element. For example, if given linked list is 1->2->3->4->5->6 then the output should be 4. 

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. 

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

Below image shows how printMiddle function works in the code :

middle-of-a-given-linked-list-in-C-and-Java1

 

Python3




# Python3 program to find middle of
# the linked list
# Node class
class Node:
   
    # Function to initialise the
    # node object
    def __init__(self, data):
    
        # Assign data
        self.data = data
 
        # Initialize next as null 
        self.next = None 
   
# Linked List class contains a
# Node object
class LinkedList:
   
    # Function to initialize head
    def __init__(self):
        self.head = None
 
    # Function to insert a new node at
    # the beginning 
    def push(self, new_data): 
        new_node = Node(new_data) 
        new_node.next = self.head 
        self.head = new_node
 
    # Print the linked list
    def printList(self):
        node = self.head
        while node:
            print(str(node.data) +
                  "->", end = "")
            node = node.next
        print("NULL")
 
    # Function that returns middle.
    def printMiddle(self):
 
        # Initialize two pointers, one will go
        # one step a time (slow), another two
        # at a time (fast)
        slow = self.head
        fast = self.head
 
        # Iterate till fast's next is null (fast
        # reaches end)
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
         
        # Return the slow's data, which would be
        # the middle element.
        print("The middle element is ", slow.data)
 
# Driver code
if __name__=='__main__':
   
    # Start with the empty list
    llist = LinkedList()
   
    for i in range(5, 0, -1):
        llist.push(i)
        llist.printList()
        llist.printMiddle()
 
# This code is contributed by Kumar Shivam (kshivi99)


Output:

5->NULL
The middle element is [5]

4->5->NULL
The middle element is [5]

3->4->5->NULL
The middle element is [4]

2->3->4->5->NULL
The middle element is [4]

1->2->3->4->5->NULL
The middle element is [3]

Time Complexity: O(n) where n is the number of nodes in the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Method 3: 
Initialize mid element as head and initialize a counter as 0. Traverse the list from head, while traversing increment the counter and change mid to mid->next whenever the counter is odd. So the mid will move only half of the total length of the list. 
Thanks to Narendra Kangralkar for suggesting this method.  

Python3




# Python program to implement
# the above approach
 
# Node class
class Node:
    
    # Function to initialise the
    # node object
    def __init__(self, data):
 
        # Assign data 
        self.data = data 
 
        # Initialize next as null
        self.next = None 
    
# Linked List class contains a
# Node object
class LinkedList:
    
    # Function to initialize head
    def __init__(self):
        self.head = None
  
    # Function to insert a new node at
    # the beginning 
    def push(self, new_data): 
        new_node = Node(new_data) 
        new_node.next = self.head 
        self.head = new_node
  
    # Print the linked list
    def printList(self):
        node = self.head
        while node:
            print(str(node.data) +
                      "->", end = "")
            node = node.next
        print("NULL")
  
    # Function to get the middle of
    #  the linked list
    def printMiddle(self):
        count = 0
        mid = self.head
        heads = self.head
   
        while(heads != None):
       
        # Update mid, when 'count'
        # is odd number
            if count & 1:
                mid = mid.next
            count += 1
            heads = heads.next
             
        # If empty list is provided
        if mid != None:
            print("The middle element is ",
                   mid.data)
  
# Driver code
if __name__=='__main__':
    
    # Start with the empty list
    llist = LinkedList()
    
    for i in range(5, 0, -1):
        llist.push(i)
        llist.printList()
        llist.printMiddle()
  
 # This code is contributed by Manisha_Ediga


Output:

5->NULL
The middle element is [5]

4->5->NULL
The middle element is [5]

3->4->5->NULL
The middle element is [4]

2->3->4->5->NULL
The middle element is [4]

1->2->3->4->5->NULL
The middle element is [3]

Time Complexity: O(n) where n is the number of nodes in the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Please refer complete article on Find the middle of a given linked list for more details!


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