Dada una lista enlazada de enteros, escriba una función para modificar la lista enlazada de modo que todos los números pares aparezcan antes que todos los números impares en la lista enlazada modificada. Además, mantén el mismo orden de los números pares e impares.
Ejemplos:
Input: 17->15->8->12->10->5->4->1->7->6->NULL Output: 8->12->10->4->6->17->15->5->1->7->NULL Input: 8->12->10->5->4->1->6->NULL Output: 8->12->10->4->6->5->1->NULL // If all numbers are even then do not change the list Input: 8->12->10->NULL Output: 8->12->10->NULL // If all numbers are odd then do not change the list Input: 1->3->5->7->NULL Output: 1->3->5->7->NULL
Método 1
La idea es obtener un puntero al último Node de la lista. Y luego recorra la lista comenzando desde el Node principal y mueva los Nodes con valores impares desde su posición actual hasta el final de la lista.
Gracias a Blunderboy por sugerir este método.
Algoritmo:
…1) Obtener puntero al último Node.
…2) Mueva todos los Nodes impares hasta el final.
……..a) Considere todos los Nodes impares antes del primer Node par y muévalos al final.
……..b) Cambie el puntero de la cabeza para que apunte al primer Node par.
……..b) Considere todos los Nodes impares después del primer Node par y muévalos al final.
C++
// C++ program to segregate even and
// odd nodes in a Linked List
#include <bits/stdc++.h>
using namespace std;
/* a node of the singly linked list */
class Node
{
public:
int data;
Node *next;
};
void segregateEvenOdd(Node **head_ref)
{
Node *end = *head_ref;
Node *prev = NULL;
Node *curr = *head_ref;
/* Get pointer to the last node */
while (end->next != NULL)
end = end->next;
Node *new_end = end;
/* Consider all odd nodes before the first
even node and move then after end */
while (curr->data % 2 != 0 && curr != end)
{
new_end->next = curr;
curr = curr->next;
new_end->next->next = NULL;
new_end = new_end->next;
}
// 10->8->17->17->15
/* Do following steps only if
there is any even node */
if (curr->data%2 == 0)
{
/* Change the head pointer to
point to first even node */
*head_ref = curr;
/* now current points to
the first even node */
while (curr != end)
{
if ( (curr->data) % 2 == 0 )
{
prev = curr;
curr = curr->next;
}
else
{
/* break the link between
prev and current */
prev->next = curr->next;
/* Make next of curr as NULL */
curr->next = NULL;
/* Move curr to end */
new_end->next = curr;
/* make curr as new end of list */
new_end = curr;
/* Update current pointer to
next of the moved node */
curr = prev->next;
}
}
}
/* We must have prev set before executing
lines following this statement */
else prev = curr;
/* If there are more than 1 odd nodes
and end of original list is odd then
move this node to end to maintain
same order of odd numbers in modified list */
if (new_end != end && (end->data) % 2 != 0)
{
prev->next = end->next;
end->next = NULL;
new_end->next = end;
}
return;
}
/* UTILITY FUNCTIONS */
/* Function to insert a node at the beginning */
void push(Node** head_ref, int new_data)
{
/* allocate node */
Node* new_node = new Node();
/* 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 point to the new node */
(*head_ref) = new_node;
}
/* Function to print nodes in a given linked list */
void printList(Node *node)
{
while (node != NULL)
{
cout << node->data <<" ";
node = node->next;
}
}
/* Driver code*/
int main()
{
/* Start with the empty list */
Node* head = NULL;
/* Let us create a sample linked list as following
0->2->4->6->8->10->11 */
push(&head, 11);
push(&head, 10);
push(&head, 8);
push(&head, 6);
push(&head, 4);
push(&head, 2);
push(&head, 0);
cout << "Original Linked list ";
printList(head);
segregateEvenOdd(&head);
cout << "\nModified Linked list ";
printList(head);
return 0;
}
// This code is contributed by rathbhupendra
C
// C program to segregate even and odd nodes in a
// Linked List
#include <stdio.h>
#include <stdlib.h>
/* a node of the singly linked list */
struct Node
{
int data;
struct Node *next;
};
void segregateEvenOdd(struct Node **head_ref)
{
struct Node *end = *head_ref;
struct Node *prev = NULL;
struct Node *curr = *head_ref;
/* Get pointer to the last node */
while (end->next != NULL)
end = end->next;
struct Node *new_end = end;
/* Consider all odd nodes before the first even node
and move then after end */
while (curr->data %2 != 0 && curr != end)
{
new_end->next = curr;
curr = curr->next;
new_end->next->next = NULL;
new_end = new_end->next;
}
// 10->8->17->17->15
/* Do following steps only if there is any even node */
if (curr->data%2 == 0)
{
/* Change the head pointer to point to first even node */
*head_ref = curr;
/* now current points to the first even node */
while (curr != end)
{
if ( (curr->data)%2 == 0 )
{
prev = curr;
curr = curr->next;
}
else
{
/* break the link between prev and current */
prev->next = curr->next;
/* Make next of curr as NULL */
curr->next = NULL;
/* Move curr to end */
new_end->next = curr;
/* make curr as new end of list */
new_end = curr;
/* Update current pointer to next of the moved node */
curr = prev->next;
}
}
}
/* We must have prev set before executing lines following this
statement */
else prev = curr;
/* If there are more than 1 odd nodes and end of original list is
odd then move this node to end to maintain same order of odd
numbers in modified list */
if (new_end!=end && (end->data)%2 != 0)
{
prev->next = end->next;
end->next = NULL;
new_end->next = end;
}
return;
}
/* UTILITY FUNCTIONS */
/* Function to insert a node at the beginning */
void push(struct Node** head_ref, int new_data)
{
/* allocate node */
struct Node* new_node =
(struct Node*) malloc(sizeof(struct Node));
/* 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 point to the new node */
(*head_ref) = new_node;
}
/* Function to print nodes in a given linked list */
void printList(struct Node *node)
{
while (node!=NULL)
{
printf("%d ", node->data);
node = node->next;
}
}
/* Driver program to test above functions*/
int main()
{
/* Start with the empty list */
struct Node* head = NULL;
/* Let us create a sample linked list as following
0->2->4->6->8->10->11 */
push(&head, 11);
push(&head, 10);
push(&head, 8);
push(&head, 6);
push(&head, 4);
push(&head, 2);
push(&head, 0);
printf("\nOriginal Linked list \n");
printList(head);
segregateEvenOdd(&head);
printf("\nModified Linked list \n");
printList(head);
return 0;
}
Java
// Java program to segregate even and odd nodes in a
// Linked List
class LinkedList
{
Node head; // head of list
/* Linked list Node*/
class Node
{
int data;
Node next;
Node(int d)
{
data = d;
next = null;
}
}
void segregateEvenOdd()
{
Node end = head;
Node prev = null;
Node curr = head;
/* Get pointer to last Node */
while (end.next != null)
end = end.next;
Node new_end = end;
// Consider all odd nodes before getting first even node
while (curr.data %2 !=0 && curr != end)
{
new_end.next = curr;
curr = curr.next;
new_end.next.next = null;
new_end = new_end.next;
}
// do following steps only if there is an even node
if (curr.data %2 ==0)
{
head = curr;
// now curr points to first even node
while (curr != end)
{
if (curr.data % 2 == 0)
{
prev = curr;
curr = curr.next;
}
else
{
/* Break the link between prev and curr*/
prev.next = curr.next;
/* Make next of curr as null */
curr.next = null;
/*Move curr to end */
new_end.next = curr;
/*Make curr as new end of list */
new_end = curr;
/*Update curr pointer */
curr = prev.next;
}
}
}
/* We have to set prev before executing rest of this code */
else prev = curr;
if (new_end != end && end.data %2 != 0)
{
prev.next = end.next;
end.next = null;
new_end.next = end;
}
}
/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);
/* 3. Make next of new Node as head */
new_node.next = head;
/* 4. Move the head to point to new Node */
head = new_node;
}
// Utility function to print a linked list
void printList()
{
Node temp = head;
while(temp != null)
{
System.out.print(temp.data+" ");
temp = temp.next;
}
System.out.println();
}
/* Driver program to test above functions */
public static void main(String args[])
{
LinkedList llist = new LinkedList();
llist.push(11);
llist.push(10);
llist.push(8);
llist.push(6);
llist.push(4);
llist.push(2);
llist.push(0);
System.out.println("Original Linked List");
llist.printList();
llist.segregateEvenOdd();
System.out.println("Modified Linked List");
llist.printList();
}
} /* This code is contributed by Rajat Mishra */
Python
# Python program to segregate even and odd nodes in a
# Linked List
head = None # head of list
# Node class
class Node:
# Function to initialise the node object
def __init__(self, data):
self.data = data # Assign data
self.next =None
def segregateEvenOdd():
global head
end = head
prev = None
curr = head
# Get pointer to last Node
while (end.next != None):
end = end.next
new_end = end
# Consider all odd nodes before getting first even node
while (curr.data % 2 !=0 and curr != end):
new_end.next = curr
curr = curr.next
new_end.next.next = None
new_end = new_end.next
# do following steps only if there is an even node
if (curr.data % 2 == 0):
head = curr
# now curr points to first even node
while (curr != end):
if (curr.data % 2 == 0):
prev = curr
curr = curr.next
else:
# Break the link between prev and curr
prev.next = curr.next
# Make next of curr as None
curr.next = None
# Move curr to end
new_end.next = curr
# Make curr as new end of list
new_end = curr
# Update curr pointer
curr = prev.next
# We have to set prev before executing rest of this code
else:
prev = curr
if (new_end != end and end.data % 2 != 0):
prev.next = end.next
end.next = None
new_end.next = end
# Given a reference (pointer to pointer) to the head
# of a list and an int, push a new node on the front
# of the list.
def push(new_data):
global head
# 1 & 2: Allocate the Node &
# Put in the data
new_node = Node(new_data)
# 3. Make next of new Node as head
new_node.next = head
# 4. Move the head to point to new Node
head = new_node
# Utility function to print a linked list
def printList():
global head
temp = head
while(temp != None):
print(temp.data, end = " ")
temp = temp.next
print(" ")
# Driver program to test above functions
push(11)
push(10)
push(8)
push(6)
push(4)
push(2)
push(0)
print("Original Linked List")
printList()
segregateEvenOdd()
print("Modified Linked List")
printList()
# This code is contributed by Arnab Kundu
C#
// C# program to segregate even and odd nodes in a
// Linked List
using System;
public class LinkedList
{
Node head; // head of list
/* Linked list Node*/
public class Node
{
public int data;
public Node next;
public Node(int d)
{
data = d;
next = null;
}
}
void segregateEvenOdd()
{
Node end = head;
Node prev = null;
Node curr = head;
/* Get pointer to last Node */
while (end.next != null)
end = end.next;
Node new_end = end;
// Consider all odd nodes before
// getting first even node
while (curr.data % 2 != 0 && curr != end)
{
new_end.next = curr;
curr = curr.next;
new_end.next.next = null;
new_end = new_end.next;
}
// do following steps only
// if there is an even node
if (curr.data % 2 == 0)
{
head = curr;
// now curr points to first even node
while (curr != end)
{
if (curr.data % 2 == 0)
{
prev = curr;
curr = curr.next;
}
else
{
/* Break the link between prev and curr*/
prev.next = curr.next;
/* Make next of curr as null */
curr.next = null;
/*Move curr to end */
new_end.next = curr;
/*Make curr as new end of list */
new_end = curr;
/*Update curr pointer */
curr = prev.next;
}
}
}
/* We have to set prev before
executing rest of this code */
else prev = curr;
if (new_end != end && end.data % 2 != 0)
{
prev.next = end.next;
end.next = null;
new_end.next = end;
}
}
/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);
/* 3. Make next of new Node as head */
new_node.next = head;
/* 4. Move the head to point to new Node */
head = new_node;
}
// Utility function to print a linked list
void printList()
{
Node temp = head;
while(temp != null)
{
Console.Write(temp.data + " ");
temp = temp.next;
}
Console.WriteLine();
}
/* Driver code */
public static void Main(String []args)
{
LinkedList llist = new LinkedList();
llist.push(11);
llist.push(10);
llist.push(8);
llist.push(6);
llist.push(4);
llist.push(2);
llist.push(0);
Console.WriteLine("Original Linked List");
llist.printList();
llist.segregateEvenOdd();
Console.WriteLine("Modified Linked List");
llist.printList();
}
}
// This code has been contributed by 29AjayKumar
Javascript
<script>
// javascript program to segregate even and odd nodes in a
// Linked List
var head; // head of list
/* Linked list Node */
class Node {
constructor(val) {
this.data = val;
this.next = null;
}
}
function segregateEvenOdd() {
var end = head;
var prev = null;
var curr = head;
/* Get pointer to last Node */
while (end.next != null)
end = end.next;
var new_end = end;
// Consider all odd nodes before getting first even node
while (curr.data % 2 != 0 && curr != end) {
new_end.next = curr;
curr = curr.next;
new_end.next.next = null;
new_end = new_end.next;
}
// do following steps only if there is an even node
if (curr.data % 2 == 0) {
head = curr;
// now curr points to first even node
while (curr != end) {
if (curr.data % 2 == 0) {
prev = curr;
curr = curr.next;
} else {
/* Break the link between prev and curr */
prev.next = curr.next;
/* Make next of curr as null */
curr.next = null;
/* Move curr to end */
new_end.next = curr;
/* Make curr as new end of list */
new_end = curr;
/* Update curr pointer */
curr = prev.next;
}
}
}
/* We have to set prev before executing rest of this code */
else
prev = curr;
if (new_end != end && end.data % 2 != 0) {
prev.next = end.next;
end.next = null;
new_end.next = end;
}
}
/*
* Given a reference (pointer to pointer) to the head of a list and an int, push
* a new node on the front of the list.
*/
function push(new_data) {
/*
* 1 & 2: Allocate the Node & Put in the data
*/
var new_node = new Node(new_data);
/* 3. Make next of new Node as head */
new_node.next = head;
/* 4. Move the head to point to new Node */
head = new_node;
}
// Utility function to print a linked list
function printList() {
var temp = head;
while (temp != null) {
document.write(temp.data + " ");
temp = temp.next;
}
document.write();
}
/* Driver program to test above functions */
push(11);
push(10);
push(8);
push(6);
push(4);
push(2);
push(0);
document.write("Original Linked List ");
printList();
document.write("<br>");
segregateEvenOdd();
document.write("Modified Linked List ");
// This code contributed by umadevi9616
</script>
Producción:
Original Linked list 0 2 4 6 8 10 11 Modified Linked list 0 2 4 6 8 10 11
Complejidad temporal: O(n)
Método 2
La idea es dividir la lista enlazada en dos: una que contenga todos los Nodes pares y otra que contenga todos los Nodes impares. Y finalmente, adjunte la lista vinculada de Nodes impares después de la lista vinculada de Nodes pares.
Para dividir la Lista vinculada, recorra la Lista vinculada original y mueva todos los Nodes impares a una Lista vinculada separada de todos los Nodes impares. Al final del ciclo, la lista original tendrá todos los Nodes pares y la lista de Nodes impares tendrá todos los Nodes impares. Para mantener el mismo orden de todos los Nodes, debemos insertar todos los Nodes impares al final de la lista de Nodes impares. Y para hacerlo en tiempo constante, debemos realizar un seguimiento del último puntero en la lista de Nodes impares.
C++
// CPP program to segregate even and odd nodes in a
// Linked List
#include <bits/stdc++.h>
using namespace std;
/* a node of the singly linked list */
struct Node {
int data;
struct Node* next;
};
// Function to segregate even and odd nodes.
void segregateEvenOdd(struct Node** head_ref)
{
// Starting node of list having even values.
Node* evenStart = NULL;
// Ending node of even values list.
Node* evenEnd = NULL;
// Starting node of odd values list.
Node* oddStart = NULL;
// Ending node of odd values list.
Node* oddEnd = NULL;
// Node to traverse the list.
Node* currNode = *head_ref;
while (currNode != NULL) {
int val = currNode->data;
// If current value is even, add it to even values
// list.
if (val % 2 == 0) {
if (evenStart == NULL) {
evenStart = currNode;
evenEnd = evenStart;
}
else {
evenEnd->next = currNode;
evenEnd = evenEnd->next;
}
}
// If current value is odd, add it to odd values
// list.
else {
if (oddStart == NULL) {
oddStart = currNode;
oddEnd = oddStart;
}
else {
oddEnd->next = currNode;
oddEnd = oddEnd->next;
}
}
// Move head pointer one step in forward direction
currNode = currNode->next;
}
// If either odd list or even list is empty, no change
// is required as all elements are either even or odd.
if (oddStart == NULL || evenStart == NULL)
return;
// Add odd list after even list.
evenEnd->next = oddStart;
oddEnd->next = NULL;
// Modify head pointer to starting of even list.
*head_ref = evenStart;
}
/* UTILITY FUNCTIONS */
/* Function to insert a node at the beginning */
void push(Node** head_ref, int new_data)
{
/* allocate node */
Node* new_node = new Node();
/* 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 point to the new node */
(*head_ref) = new_node;
}
/* Function to print nodes in a given linked list */
void printList(Node* node)
{
while (node != NULL) {
cout << node->data << " ";
node = node->next;
}
}
/* Driver program to test above functions*/
int main()
{
/* Start with the empty list */
Node* head = NULL;
/* Let us create a sample linked list as following
0->1->4->6->9->10->11 */
push(&head, 11);
push(&head, 10);
push(&head, 9);
push(&head, 6);
push(&head, 4);
push(&head, 1);
push(&head, 0);
cout << "Original Linked list" << endl;
printList(head);
segregateEvenOdd(&head);
cout << endl << "Modified Linked list " << endl;
printList(head);
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
C
// C program to segregate even and odd nodes in a
// Linked List
#include <stdio.h>
#include <stdlib.h>
/* a node of the singly linked list */
typedef struct Node {
int data;
struct Node* next;
} Node;
// Function to segregate even and odd nodes.
void segregateEvenOdd(Node** head_ref)
{
// Starting node of list having
// even values.
Node* evenStart = NULL;
// Ending node of even values list.
Node* evenEnd = NULL;
// Starting node of odd values list.
Node* oddStart = NULL;
// Ending node of odd values list.
Node* oddEnd = NULL;
// Node to traverse the list.
Node* currNode = *head_ref;
while (currNode != NULL) {
int val = currNode->data;
// If current value is even, add it to even values
// list.
if (val % 2 == 0) {
if (evenStart == NULL) {
evenStart = currNode;
evenEnd = evenStart;
}
else {
evenEnd->next = currNode,
evenEnd = evenEnd->next;
}
}
// If current value is odd, add it to odd values
// list.
else {
if (oddStart == NULL) {
oddStart = currNode;
oddEnd = oddStart;
}
else {
oddEnd->next = currNode;
oddEnd = oddEnd->next;
}
}
// Move head pointer one step in forward direction
currNode = currNode->next;
}
// If either odd list or even list is empty, no change
// is required as all elements are either even or odd.
if (oddStart == NULL || evenStart == NULL)
return;
// Add odd list after even list.
evenEnd->next = oddStart;
oddEnd->next = NULL;
// Modify head pointer to
// starting of even list.
*head_ref = evenStart;
}
/* UTILITY FUNCTIONS */
/* Function to insert a node at the beginning */
void push(Node** head_ref, int new_data)
{
/* allocate node */
Node* new_node = (Node*)malloc(sizeof(Node));
/* 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 point to the new node */
(*head_ref) = new_node;
}
/* Function to print nodes in a given linked list */
void printList(Node* node)
{
while (node != NULL) {
printf("%d ", node->data);
node = node->next;
}
}
/* Driver program to test above functions*/
int main()
{
/* Start with the empty list */
struct Node* head = NULL;
/* Let us create a sample linked list as following
0->1->4->6->9->10->11 */
push(&head, 11);
push(&head, 10);
push(&head, 9);
push(&head, 6);
push(&head, 4);
push(&head, 1);
push(&head, 0);
printf("\nOriginal Linked list \n");
printList(head);
segregateEvenOdd(&head);
printf("\nModified Linked list \n");
printList(head);
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
Java
// Java program to segregate even and odd nodes in a
// Linked List
import java.io.*;
class LinkedList {
Node head; // head of list
/* Linked list Node*/
class Node
{
int data;
Node next;
Node(int d)
{
data = d;
next = null;
}
}
public void segregateEvenOdd() {
Node evenStart = null;
Node evenEnd = null;
Node oddStart = null;
Node oddEnd = null;
Node currentNode = head;
while(currentNode != null) {
int element = currentNode.data;
if(element % 2 == 0) {
if(evenStart == null) {
evenStart = currentNode;
evenEnd = evenStart;
} else {
evenEnd.next = currentNode;
evenEnd = evenEnd.next;
}
} else {
if(oddStart == null) {
oddStart = currentNode;
oddEnd = oddStart;
} else {
oddEnd.next = currentNode;
oddEnd = oddEnd.next;
}
}
// Move head pointer one step in forward direction
currentNode = currentNode.next;
}
if(oddStart == null || evenStart == null) {
return;
}
evenEnd.next = oddStart;
oddEnd.next = null;
head=evenStart;
}
/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);
/* 3. Make next of new Node as head */
new_node.next = head;
/* 4. Move the head to point to new Node */
head = new_node;
}
// Utility function to print a linked list
void printList()
{
Node temp = head;
while(temp != null)
{
System.out.print(temp.data+" ");
temp = temp.next;
}
System.out.println();
}
/* Driver program to test above functions */
public static void main(String args[])
{
LinkedList llist = new LinkedList();
llist.push(11);
llist.push(10);
llist.push(9);
llist.push(6);
llist.push(4);
llist.push(1);
llist.push(0);
System.out.println("Original Linked List");
llist.printList();
llist.segregateEvenOdd();
System.out.println("Modified Linked List");
llist.printList();
}
}
C#
// C# program to segregate even and odd nodes in a
// Linked List
using System;
public class LinkedList
{
Node head; // head of list
/* Linked list Node*/
public class Node
{
public int data;
public Node next;
public Node(int d)
{
data = d;
next = null;
}
}
public void segregateEvenOdd()
{
Node evenStart = null;
Node evenEnd = null;
Node oddStart = null;
Node oddEnd = null;
Node currentNode = head;
while(currentNode != null)
{
int element = currentNode.data;
if(element % 2 == 0)
{
if(evenStart == null)
{
evenStart = currentNode;
evenEnd = evenStart;
}
else
{
evenEnd.next = currentNode;
evenEnd = evenEnd.next;
}
}
else
{
if(oddStart == null)
{
oddStart = currentNode;
oddEnd = oddStart;
}
else
{
oddEnd.next = currentNode;
oddEnd = oddEnd.next;
}
}
// Move head pointer one step in forward direction
currentNode = currentNode.next;
}
if(oddStart == null || evenStart == null)
{
return;
}
evenEnd.next = oddStart;
oddEnd.next = null;
head=evenStart;
}
/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
void push(int new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
Node new_node = new Node(new_data);
/* 3. Make next of new Node as head */
new_node.next = head;
/* 4. Move the head to point to new Node */
head = new_node;
}
// Utility function to print a linked list
void printList()
{
Node temp = head;
while(temp != null)
{
Console.Write(temp.data+" ");
temp = temp.next;
}
Console.WriteLine();
}
/* Driver code */
public static void Main()
{
LinkedList llist = new LinkedList();
llist.push(11);
llist.push(10);
llist.push(9);
llist.push(6);
llist.push(4);
llist.push(1);
llist.push(0);
Console.WriteLine("Original Linked List");
llist.printList();
llist.segregateEvenOdd();
Console.WriteLine("Modified Linked List");
llist.printList();
}
}
/* This code contributed by PrinciRaj1992 */
Python3
# Python3 program to segregate even and odd nodes in a
# Linked List
head = None # head of list
# Node class
class Node:
# Function to initialise the node object
def __init__(self, data):
self.data = data # Assign data
self.next =None
# Function to segregate even and odd nodes.
def segregateEvenOdd():
global head
# Starting node of list having
# even values.
evenStart = None
# Ending node of even values list.
evenEnd = None
# Starting node of odd values list.
oddStart = None
# Ending node of odd values list.
oddEnd = None
# Node to traverse the list.
currNode = head
while(currNode != None):
val = currNode.data
# If current value is even, add
# it to even values list.
if(val % 2 == 0):
if(evenStart == None):
evenStart = currNode
evenEnd = evenStart
else:
evenEnd . next = currNode
evenEnd = evenEnd . next
# If current value is odd, add
# it to odd values list.
else:
if(oddStart == None):
oddStart = currNode
oddEnd = oddStart
else:
oddEnd . next = currNode
oddEnd = oddEnd . next
# Move head pointer one step in
# forward direction
currNode = currNode . next
# If either odd list or even list is empty,
# no change is required as all elements
# are either even or odd.
if(oddStart == None or evenStart == None):
return
# Add odd list after even list.
evenEnd . next = oddStart
oddEnd . next = None
# Modify head pointer to
# starting of even list.
head = evenStart
''' UTILITY FUNCTIONS '''
''' Function to insert a node at the beginning '''
def push(new_data):
global head
# 1 & 2: Allocate the Node &
# Put in the data
new_node = Node(new_data)
# 3. Make next of new Node as head
new_node.next = head
# 4. Move the head to point to new Node
head = new_node
''' Function to print nodes in a given linked list '''
def printList():
global head
node = head
while (node != None):
print(node.data, end = " ")
node = node.next
print()
''' Driver program to test above functions'''
''' Let us create a sample linked list as following
0.1.4.6.9.10.11 '''
push(11)
push(10)
push(9)
push(6)
push(4)
push(1)
push(0)
print("Original Linked list")
printList()
segregateEvenOdd()
print("Modified Linked list")
printList()
# This code is contributed by shubhamsingh10.
Javascript
<script>
// JavaScript program to segregate
// even and odd nodes in a
// Linked List
var head; // head of list
/* Linked list Node*/
class Node {
constructor(val) {
this.data = val;
this.next = null;
}
}
function segregateEvenOdd() {
var evenStart = null;
var evenEnd = null;
var oddStart = null;
var oddEnd = null;
var currentNode = head;
while(currentNode != null) {
var element = currentNode.data;
if(element % 2 == 0) {
if(evenStart == null) {
evenStart = currentNode;
evenEnd = evenStart;
} else {
evenEnd.next = currentNode;
evenEnd = evenEnd.next;
}
} else {
if(oddStart == null) {
oddStart = currentNode;
oddEnd = oddStart;
} else {
oddEnd.next = currentNode;
oddEnd = oddEnd.next;
}
}
// Move head pointer one
// step in forward direction
currentNode = currentNode.next;
}
if(oddStart == null || evenStart == null) {
return;
}
evenEnd.next = oddStart;
oddEnd.next = null;
head=evenStart;
}
/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
function push(new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
var new_node = new Node(new_data);
/* 3. Make next of new Node as head */
new_node.next = head;
/* 4. Move the head to point to new Node */
head = new_node;
}
// Utility function to print a linked list
function printList()
{
var temp = head;
while(temp != null)
{
document.write(temp.data+" ");
temp = temp.next;
}
document.write("<br/>");
}
/* Driver program to test above functions */
push(11);
push(10);
push(9);
push(6);
push(4);
push(1);
push(0);
document.write("Original Linked List<br/>");
printList();
segregateEvenOdd();
document.write("Modified Linked List<br/>");
printList();
// This code is contributed by todaysgaurav
</script>
Producción:
Original Linked List 0 1 4 6 9 10 11 Modified Linked List 0 4 6 10 1 9 11
Complejidad de tiempo: O(n)
Escriba comentarios si encuentra que el código/algoritmo anterior es incorrecto o encuentra otras formas de resolver el mismo problema.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA