Python Program For Finding The Middle Element Of A Given Linked List
Last Updated :
22 Jun, 2022
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 :
Python3
class Node:
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
def printList( self ):
node = self .head
while node:
print ( str (node.data) +
"->" , end = "")
node = node. next
print ( "NULL" )
def printMiddle( self ):
slow = self .head
fast = self .head
while fast and fast. next :
slow = slow. next
fast = fast. next . next
print ( "The middle element is " , slow.data)
if __name__ = = '__main__' :
llist = LinkedList()
for i in range ( 5 , 0 , - 1 ):
llist.push(i)
llist.printList()
llist.printMiddle()
|
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
class Node:
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
def printList( self ):
node = self .head
while node:
print ( str (node.data) +
"->" , end = "")
node = node. next
print ( "NULL" )
def printMiddle( self ):
count = 0
mid = self .head
heads = self .head
while (heads ! = None ):
if count & 1 :
mid = mid. next
count + = 1
heads = heads. next
if mid ! = None :
print ( "The middle element is " ,
mid.data)
if __name__ = = '__main__' :
llist = LinkedList()
for i in range ( 5 , 0 , - 1 ):
llist.push(i)
llist.printList()
llist.printMiddle()
|
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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