Dada una lista enlazada, escriba una función para invertir cada k Node alternativo (donde k es una entrada a la función) de manera eficiente. Da la complejidad de tu algoritmo.
Ejemplo:
Inputs: 1->2->3->4->5->6->7->8->9->NULL and k = 3 Output: 3->2->1->4->5->6->9->8->7->NULL.
Método 1 (procesar 2k Nodes y llamar recursivamente al resto de la lista):
este método es básicamente una extensión del método discutido en esta publicación.
kAltReverse(struct node *head, int k)
1) Reverse first k nodes.
2) In the modified list head points to the kth node. So change next
of head to (k+1)th node
3) Move the current pointer to skip next k nodes.
4) Call the kAltReverse() recursively for rest of the n - 2k nodes.
5) Return new head of the list.
Python3
# Python3 program to reverse alternate
# k nodes in a linked list
import math
# Link list node
class Node:
def __init__(self, data):
self.data = data
self.next = None
# Reverses alternate k nodes and
#returns the pointer to the new
# head node
def kAltReverse(head, k) :
current = head
next = None
prev = None
count = 0
# 1) reverse first k nodes of the
# linked list
while (current != None and
count < k) :
next = current.next
current.next = prev
prev = current
current = next
count = count + 1;
# 2) Now head pos to the kth node.
# So change next of head to (k+1)th node
if(head != None):
head.next = current
# 3) We do not want to reverse next k
# nodes. So move the current
# pointer to skip next k nodes
count = 0
while(count < k - 1 and
current != None ):
current = current.next
count = count + 1
# 4) Recursively call for the list
# starting from current.next. And make
# rest of the list as next of first node
if(current != None):
current.next =
kAltReverse(current.next, k)
# 5) prev is the new head of the
# input list
return prev
# UTILITY FUNCTIONS
# Function to push a node
def push(head_ref, new_data):
# Allocate node
new_node = Node(new_data)
# Put in the data
# new_node.data = new_data
# Link the old list off the
# new node
new_node.next = head_ref
# Move the head to po to the
# new node
head_ref = new_node
return head_ref
# Function to print linked list
def printList(node):
count = 0
while(node != None):
print(node.data, end = " ")
node = node.next
count = count + 1
# Driver code
if __name__=='__main__':
# Start with the empty list
head = None
# Create a list
# 1.2.3.4.5...... .20
for i in range(20, 0, -1):
head = push(head, i)
print("Given linked list ")
printList(head)
head = kAltReverse(head, 3)
print("Modified Linked list")
printList(head)
# This code is contributed by Srathore
Producción:
Given linked list 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Modified Linked list 3 2 1 4 5 6 9 8 7 10 11 12 15 14 13 16 17 18 20 19
Complejidad de tiempo: O(n)
Método 2 (procesar k Nodes y llamar recursivamente al resto de la lista):
el método 1 invierte el primer k Node y luego mueve el puntero a k Nodes adelante. Entonces, el método 1 usa dos bucles while y procesa 2k Nodes en una llamada recursiva.
Este método procesa solo k Nodes en una llamada recursiva. Utiliza un tercer parámetro bool b que decide si invertir los k elementos o simplemente mover el puntero.
_kAltReverse(struct node *head, int k, bool b)
1) If b is true, then reverse first k nodes.
2) If b is false, then move the pointer k nodes ahead.
3) Call the kAltReverse() recursively for rest of the n - k nodes and link
rest of the modified list with end of first k nodes.
4) Return new head of the list.
Python3
# Python code for above algorithm
# Link list node
class node:
def __init__(self, data):
self.data = data
self.next = next
# Function to insert a node at the
# beginning of the linked list
def push(head_ref, new_data):
# Allocate node
new_node = node(0)
# Put in the data
new_node.data = new_data
# Link the old list to the
# new node
new_node.next = (head_ref)
# Move the head to point to the
# new node
(head_ref) = new_node
return head_ref
""" Alternatively reverses the given
linked list in groups of given
size k. """
def kAltReverse(head, k) :
return _kAltReverse(head, k, True)
""" Helper function for kAltReverse().
It reverses k nodes of the list only
if the third parameter b is passed as
True, otherwise moves the pointer k
nodes ahead and recursively call itself """
def _kAltReverse(Node, k, b) :
if(Node == None) :
return None
count = 1
prev = None
current = Node
next = None
""" The loop serves two purposes
1) If b is True,
then it reverses the k nodes
2) If b is false,
then it moves the current pointer """
while(current != None and count <= k) :
next = current.next
""" Reverse the nodes only if b
is True """
if(b == True) :
current.next = prev
prev = current
current = next
count = count + 1
""" 3) If b is True, then node is the
kth node. So attach rest of the list
after node.
4) After attaching, return the new
head """
if(b == True) :
Node.next =
_kAltReverse(current,
k, not b)
return prev
else :
""" If b is not True, then attach
rest of the list after prev.
So attach rest of the list
after prev """
prev.next = _kAltReverse(current,
k, not b)
return Node
""" Function to print linked list """
def printList(node) :
count = 0
while(node != None) :
print( node.data, end = " ")
node = node.next
count = count + 1
# Driver Code
# Start with the empty list
head = None
i = 20
# Create a list
# 1->2->3->4->5...... ->20
while(i > 0 ):
head = push(head, i)
i = i - 1
print("Given linked list ")
printList(head)
head = kAltReverse(head, 3)
print("Modified Linked list ")
printList(head)
# This code is contributed by Arnab Kundu
Producción:
Given linked list 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Modified Linked list 3 2 1 4 5 6 9 8 7 10 11 12 15 14 13 16 17 18 20 19
Complejidad de tiempo: O(n)
¡ Consulte el artículo completo sobre los Nodes K alternativos inversos en una lista enlazada individual para obtener más detalles!
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA