Given a linked list, the task is to make a function which checks whether the length of the linked list is even or odd.
Examples:
Input : 1->2->3->4->NULL Output : Even Input : 1->2->3->4->5->NULL Output : Odd
Method 1: Count the codes linearly
Traverse the entire Linked List and keep counting the number of nodes. As soon as the loop is finished, we can check if the count is even or odd. You may try it yourself.
Method 2: Stepping 2 nodes at a time
Approach:
1. Take a pointer and move that pointer two nodes at a time 2. At the end, if the pointer is NULL then length is Even, else Odd.
# Python program to check length # of a given linklist # Defining structure class Node:
def __init__( self , d):
self .data = d
self . next = None
self .head = None
# Function to check the length
# of linklist
def LinkedListLength( self ):
while ( self .head ! = None and
self .head. next ! = None ):
self .head = self .head. next . next
if ( self .head = = None ):
return 0
return 1
# Push function
def push( self , info):
# Allocating node
node = Node(info)
# Next of new node to head
node. next = ( self .head)
# head points to new node
( self .head) = node
# Driver code head = Node( 0 )
# Adding elements to Linked List head.push( 4 )
head.push( 5 )
head.push( 7 )
head.push( 2 )
head.push( 9 )
head.push( 6 )
head.push( 1 )
head.push( 2 )
head.push( 0 )
head.push( 5 )
head.push( 5 )
check = head.LinkedListLength()
# Checking for length of # linklist if (check = = 0 ):
print ( "Even" )
else :
print ( "Odd" )
# This code is contributed by Prerna saini |
Odd
Time Complexity: O(n)
Space Complexity: O(1)
Method: Using recursion
class Node:
def __init__( self , d):
self .data = d
self . next = None
self .head = None
# Function to check the length
# of linklist
def LinkedListLength( self , current = None ):
# if current is None:
#current = self.head
if current is None :
return 0
return 1 + self .LinkedListLength(current. next )
# Push function
def push( self , info):
# Allocating node
node = Node(info)
# Next of new node to head
node. next = ( self .head)
# head points to new node
( self .head) = node
# Driver code head = Node( 0 )
# Adding elements to Linked List head.push( 4 )
head.push( 5 )
head.push( 7 )
head.push( 2 )
head.push( 9 )
head.push( 6 )
head.push( 1 )
head.push( 2 )
head.push( 0 )
head.push( 5 )
length = head.LinkedListLength()
# Checking for length of # linklist if length % 2 = = 0 :
print ( "Even" )
else :
print ( "Odd" )
# This code is contributed by Vinay Pinjala.
|
Even
Time Complexity: O(n)
Auxiliary Space: O(n)
Please refer complete article on Check whether the length of given linked list is Even or Odd for more details!
Method: Using Two pointer approach:
# Define the Node class to create nodes for the linked list class Node:
def __init__( self , data):
self .data = data
self . next = None
# Define the LinkedList class to create and manage linked lists class LinkedList:
def __init__( self ):
self .head = None
# Function to determine the length of the linked list
def length( self ):
slow_ptr = self .head # Initialize the slow pointer to the head of the list
fast_ptr = self .head # Initialize the fast pointer to the head of the list
# Move the slow pointer one step at a time
# Move the fast pointer two steps at a time
while fast_ptr is not None and fast_ptr. next is not None :
slow_ptr = slow_ptr. next
fast_ptr = fast_ptr. next . next
# If the fast pointer reaches the end of the list, the length is even
if fast_ptr is None :
return "Even"
# If the fast pointer reaches the last node of the list, the length is odd
else :
return "Odd"
# Initialize a linked list and add some nodes to it linked_list = LinkedList()
linked_list.head = Node( 1 )
linked_list.head. next = Node( 2 )
linked_list.head. next . next = Node( 3 )
linked_list.head. next . next . next = Node( 4 )
# Determine the length of the linked list length = linked_list.length()
print ( "Length:" , length)
#This code is contributed by Jyothi pinjala. |
Length: Even
Time Complexity: O(n)
Auxiliary Space: O(n)