Given pointer to the head node of a linked list, the task is to reverse the linked list. We need to reverse the list by changing links between nodes.
Examples:
Input : Head of following linked list
1->2->3->4->NULL
Output : Linked list should be changed to,
4->3->2->1->NULL
Input : Head of following linked list
1->2->3->4->5->NULL
Output : Linked list should be changed to,
5->4->3->2->1->NULL
Input : NULL
Output : NULL
Input : 1->NULL
Output : 1->NULL
Iterative Method
Python3
class Node:
def __init__( self , data):
self .data = data
self . next = None
class LinkedList:
def __init__( self ):
self .head = None
def reverse( self ):
prev = None
current = self .head
while (current is not None ):
next = current. next
current. next = prev
prev = current
current = next
self .head = prev
def push( self , new_data):
new_node = Node(new_data)
new_node. next = self .head
self .head = new_node
def printList( self ):
temp = self .head
while (temp):
print (temp.data,end = " " )
temp = temp. next
llist = LinkedList()
llist.push( 20 )
llist.push( 4 )
llist.push( 15 )
llist.push( 85 )
print ( "Given Linked List" )
llist.printList()
llist.reverse()
print ( "\nReversed Linked List" )
llist.printList()
|
Output:
Given Linked List
85 15 4 20
Reversed Linked List
20 4 15 85
Time Complexity: O(N)
Auxiliary Space: O(1)
A Simpler and Tail Recursive Method
Python3
class Node:
def __init__( self , data):
self .data = data
self . next = None
class LinkedList:
def __init__( self ):
self .head = None
def reverseUtil( self , curr, prev):
if curr. next is None :
self .head = curr
curr. next = prev
return
next = curr. next
curr. next = prev
self .reverseUtil( next , curr)
def reverse( self ):
if self .head is None :
return
self .reverseUtil( 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 ):
temp = self .head
while (temp):
print (temp.data,end = " " )
temp = temp. next
llist = LinkedList()
llist.push( 8 )
llist.push( 7 )
llist.push( 6 )
llist.push( 5 )
llist.push( 4 )
llist.push( 3 )
llist.push( 2 )
llist.push( 1 )
print ( "Given linked list" )
llist.printList()
llist.reverse()
print ( "\nReverse linked list" )
llist.printList()
|
Output
Given linked list
1 2 3 4 5 6 7 8
Reverse linked list
8 7 6 5 4 3 2 1
Time Complexity: O(N)
Auxiliary Space: O(1)
Please refer complete article on Reverse a linked list for more details!
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Last Updated :
10 Jan, 2023
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