A menudo se prefiere la ordenación por combinación para ordenar una lista vinculada. El lento rendimiento de acceso aleatorio de una lista enlazada hace que algunos otros algoritmos (como la ordenación rápida) funcionen mal y otros (como la ordenación heap) sean completamente imposibles.
Deje que head sea el primer Node de la lista enlazada que se ordenará y headRef sea el puntero a head. Tenga en cuenta que necesitamos una referencia al encabezado en MergeSort() ya que la implementación a continuación cambia los siguientes enlaces para ordenar las listas vinculadas (no los datos en los Nodes), por lo que el Node principal debe cambiarse si los datos en el encabezado original no son los valor más pequeño en la lista enlazada.
MergeSort(headRef)
1) If the head is NULL or there is only one element in the Linked List
then return.
2) Else divide the linked list into two halves.
FrontBackSplit(head, &a, &b); /* a and b are two halves */
3) Sort the two halves a and b.
MergeSort(a);
MergeSort(b);
4) Merge the sorted a and b (using SortedMerge() discussed here)
and update the head pointer using headRef.
*headRef = SortedMerge(a, b);
C++
// C++ code for linked list merged sort
#include <bits/stdc++.h>
using namespace std;
/* Link list node */
class Node {
public:
int data;
Node* next;
};
/* function prototypes */
Node* SortedMerge(Node* a, Node* b);
void FrontBackSplit(Node* source,
Node** frontRef, Node** backRef);
/* sorts the linked list by changing next pointers (not data) */
void MergeSort(Node** headRef)
{
Node* head = *headRef;
Node* a;
Node* b;
/* Base case -- length 0 or 1 */
if ((head == NULL) || (head->next == NULL)) {
return;
}
/* Split head into 'a' and 'b' sublists */
FrontBackSplit(head, &a, &b);
/* Recursively sort the sublists */
MergeSort(&a);
MergeSort(&b);
/* answer = merge the two sorted lists together */
*headRef = SortedMerge(a, b);
}
/* See https:// www.geeksforgeeks.org/?p=3622 for details of this
function */
Node* SortedMerge(Node* a, Node* b)
{
Node* result = NULL;
/* Base cases */
if (a == NULL)
return (b);
else if (b == NULL)
return (a);
/* Pick either a or b, and recur */
if (a->data <= b->data) {
result = a;
result->next = SortedMerge(a->next, b);
}
else {
result = b;
result->next = SortedMerge(a, b->next);
}
return (result);
}
/* UTILITY FUNCTIONS */
/* Split the nodes of the given list into front and back halves,
and return the two lists using the reference parameters.
If the length is odd, the extra node should go in the front list.
Uses the fast/slow pointer strategy. */
void FrontBackSplit(Node* source,
Node** frontRef, Node** backRef)
{
Node* fast;
Node* slow;
slow = source;
fast = source->next;
/* Advance 'fast' two nodes, and advance 'slow' one node */
while (fast != NULL) {
fast = fast->next;
if (fast != NULL) {
slow = slow->next;
fast = fast->next;
}
}
/* 'slow' is before the midpoint in the list, so split it in two
at that point. */
*frontRef = source;
*backRef = slow->next;
slow->next = NULL;
}
/* Function to print nodes in a given linked list */
void printList(Node* node)
{
while (node != NULL) {
cout << node->data << " ";
node = node->next;
}
}
/* Function to insert a node at the beginning of the linked list */
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;
}
/* Driver program to test above functions*/
int main()
{
/* Start with the empty list */
Node* res = NULL;
Node* a = NULL;
/* Let us create a unsorted linked lists to test the functions
Created lists shall be a: 2->3->20->5->10->15 */
push(&a, 15);
push(&a, 10);
push(&a, 5);
push(&a, 20);
push(&a, 3);
push(&a, 2);
/* Sort the above created Linked List */
MergeSort(&a);
cout << "Sorted Linked List is: \n";
printList(a);
return 0;
}
// This is code is contributed by rathbhupendra
C
// C code for linked list merged sort
#include <stdio.h>
#include <stdlib.h>
/* Link list node */
struct Node {
int data;
struct Node* next;
};
/* function prototypes */
struct Node* SortedMerge(struct Node* a, struct Node* b);
void FrontBackSplit(struct Node* source,
struct Node** frontRef, struct Node** backRef);
/* sorts the linked list by changing next pointers (not data) */
void MergeSort(struct Node** headRef)
{
struct Node* head = *headRef;
struct Node* a;
struct Node* b;
/* Base case -- length 0 or 1 */
if ((head == NULL) || (head->next == NULL)) {
return;
}
/* Split head into 'a' and 'b' sublists */
FrontBackSplit(head, &a, &b);
/* Recursively sort the sublists */
MergeSort(&a);
MergeSort(&b);
/* answer = merge the two sorted lists together */
*headRef = SortedMerge(a, b);
}
/* See https:// www.geeksforgeeks.org/?p=3622 for details of this
function */
struct Node* SortedMerge(struct Node* a, struct Node* b)
{
struct Node* result = NULL;
/* Base cases */
if (a == NULL)
return (b);
else if (b == NULL)
return (a);
/* Pick either a or b, and recur */
if (a->data <= b->data) {
result = a;
result->next = SortedMerge(a->next, b);
}
else {
result = b;
result->next = SortedMerge(a, b->next);
}
return (result);
}
/* UTILITY FUNCTIONS */
/* Split the nodes of the given list into front and back halves,
and return the two lists using the reference parameters.
If the length is odd, the extra node should go in the front list.
Uses the fast/slow pointer strategy. */
void FrontBackSplit(struct Node* source,
struct Node** frontRef, struct Node** backRef)
{
struct Node* fast;
struct Node* slow;
slow = source;
fast = source->next;
/* Advance 'fast' two nodes, and advance 'slow' one node */
while (fast != NULL) {
fast = fast->next;
if (fast != NULL) {
slow = slow->next;
fast = fast->next;
}
}
/* 'slow' is before the midpoint in the list, so split it in two
at that point. */
*frontRef = source;
*backRef = slow->next;
slow->next = NULL;
}
/* Function to print nodes in a given linked list */
void printList(struct Node* node)
{
while (node != NULL) {
printf("%d ", node->data);
node = node->next;
}
}
/* Function to insert a node at the beginning of the linked list */
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;
}
/* Driver program to test above functions*/
int main()
{
/* Start with the empty list */
struct Node* res = NULL;
struct Node* a = NULL;
/* Let us create a unsorted linked lists to test the functions
Created lists shall be a: 2->3->20->5->10->15 */
push(&a, 15);
push(&a, 10);
push(&a, 5);
push(&a, 20);
push(&a, 3);
push(&a, 2);
/* Sort the above created Linked List */
MergeSort(&a);
printf("Sorted Linked List is: \n");
printList(a);
getchar();
return 0;
}
Java
// Java program to illustrate merge sorted
// of linkedList
public class linkedList {
node head = null;
// node a, b;
static class node {
int val;
node next;
public node(int val)
{
this.val = val;
}
}
node sortedMerge(node a, node b)
{
node result = null;
/* Base cases */
if (a == null)
return b;
if (b == null)
return a;
/* Pick either a or b, and recur */
if (a.val <= b.val) {
result = a;
result.next = sortedMerge(a.next, b);
}
else {
result = b;
result.next = sortedMerge(a, b.next);
}
return result;
}
node mergeSort(node h)
{
// Base case : if head is null
if (h == null || h.next == null) {
return h;
}
// get the middle of the list
node middle = getMiddle(h);
node nextofmiddle = middle.next;
// set the next of middle node to null
middle.next = null;
// Apply mergeSort on left list
node left = mergeSort(h);
// Apply mergeSort on right list
node right = mergeSort(nextofmiddle);
// Merge the left and right lists
node sortedlist = sortedMerge(left, right);
return sortedlist;
}
// Utility function to get the middle of the linked list
public static node getMiddle(node head)
{
if (head == null)
return head;
node slow = head, fast = head;
while (fast.next != null && fast.next.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
void push(int new_data)
{
/* allocate node */
node new_node = new node(new_data);
/* link the old list off the new node */
new_node.next = head;
/* move the head to point to the new node */
head = new_node;
}
// Utility function to print the linked list
void printList(node headref)
{
while (headref != null) {
System.out.print(headref.val + " ");
headref = headref.next;
}
}
public static void main(String[] args)
{
linkedList li = new linkedList();
/*
* Let us create a unsorted linked list to test the functions
* created. The list shall be a: 2->3->20->5->10->15
*/
li.push(15);
li.push(10);
li.push(5);
li.push(20);
li.push(3);
li.push(2);
// Apply merge Sort
li.head = li.mergeSort(li.head);
System.out.print("\n Sorted Linked List is: \n");
li.printList(li.head);
}
}
// This code is contributed by Rishabh Mahrsee
Python3
# Python3 program to merge sort of linked list
# create Node using class Node.
class Node:
def __init__(self, data):
self.data = data
self.next = None
class LinkedList:
def __init__(self):
self.head = None
# push new value to linked list
# using append method
def append(self, new_value):
# Allocate new node
new_node = Node(new_value)
# if head is None, initialize it to new node
if self.head is None:
self.head = new_node
return
curr_node = self.head
while curr_node.next is not None:
curr_node = curr_node.next
# Append the new node at the end
# of the linked list
curr_node.next = new_node
def sortedMerge(self, a, b):
result = None
# Base cases
if a == None:
return b
if b == None:
return a
# pick either a or b and recur..
if a.data <= b.data:
result = a
result.next = self.sortedMerge(a.next, b)
else:
result = b
result.next = self.sortedMerge(a, b.next)
return result
def mergeSort(self, h):
# Base case if head is None
if h == None or h.next == None:
return h
# get the middle of the list
middle = self.getMiddle(h)
nexttomiddle = middle.next
# set the next of middle node to None
middle.next = None
# Apply mergeSort on left list
left = self.mergeSort(h)
# Apply mergeSort on right list
right = self.mergeSort(nexttomiddle)
# Merge the left and right lists
sortedlist = self.sortedMerge(left, right)
return sortedlist
# Utility function to get the middle
# of the linked list
def getMiddle(self, head):
if (head == None):
return head
slow = head
fast = head
while (fast.next != None and
fast.next.next != None):
slow = slow.next
fast = fast.next.next
return slow
# Utility function to print the linked list
def printList(head):
if head is None:
print(' ')
return
curr_node = head
while curr_node:
print(curr_node.data, end = " ")
curr_node = curr_node.next
print(' ')
# Driver Code
if __name__ == '__main__':
li = LinkedList()
# Let us create a unsorted linked list
# to test the functions created.
# The list shall be a: 2->3->20->5->10->15
li.append(15)
li.append(10)
li.append(5)
li.append(20)
li.append(3)
li.append(2)
# Apply merge Sort
li.head = li.mergeSort(li.head)
print ("Sorted Linked List is:")
printList(li.head)
# This code is contributed by Vikas Chitturi
C#
// C# program to illustrate merge sorted
// of linkedList
using System;
public class linkedList {
node head = null;
// node a, b;
public class node {
public int val;
public node next;
public node(int val)
{
this.val = val;
}
}
node sortedMerge(node a, node b)
{
node result = null;
/* Base cases */
if (a == null)
return b;
if (b == null)
return a;
/* Pick either a or b, and recur */
if (a.val <= b.val) {
result = a;
result.next = sortedMerge(a.next, b);
}
else {
result = b;
result.next = sortedMerge(a, b.next);
}
return result;
}
node mergeSort(node h)
{
// Base case : if head is null
if (h == null || h.next == null) {
return h;
}
// get the middle of the list
node middle = getMiddle(h);
node nextofmiddle = middle.next;
// set the next of middle node to null
middle.next = null;
// Apply mergeSort on left list
node left = mergeSort(h);
// Apply mergeSort on right list
node right = mergeSort(nextofmiddle);
// Merge the left and right lists
node sortedlist = sortedMerge(left, right);
return sortedlist;
}
// Utility function to get the
// middle of the linked list
node getMiddle(node h)
{
// Base case
if (h == null)
return h;
node fastptr = h.next;
node slowptr = h;
// Move fastptr by two and slow ptr by one
// Finally slowptr will point to middle node
while (fastptr != null) {
fastptr = fastptr.next;
if (fastptr != null) {
slowptr = slowptr.next;
fastptr = fastptr.next;
}
}
return slowptr;
}
void push(int new_data)
{
/* allocate node */
node new_node = new node(new_data);
/* link the old list off the new node */
new_node.next = head;
/* move the head to point to the new node */
head = new_node;
}
// Utility function to print the linked list
void printList(node headref)
{
while (headref != null) {
Console.Write(headref.val + " ");
headref = headref.next;
}
}
// Driver code
public static void Main(String[] args)
{
linkedList li = new linkedList();
/*
* Let us create a unsorted linked list to test the functions
* created. The list shall be a: 2->3->20->5->10->15
*/
li.push(15);
li.push(10);
li.push(5);
li.push(20);
li.push(3);
li.push(2);
// Apply merge Sort
li.head = li.mergeSort(li.head);
Console.Write("\n Sorted Linked List is: \n");
li.printList(li.head);
}
}
// This code is contributed by Arnab Kundu
Javascript
<script>
// Javascript program to
// illustrate merge sorted
// of linkedList
var head = null;
// node a, b;
class node {
constructor(val) {
this.val = val;
this.next = null;
}
}
function sortedMerge( a, b)
{
var result = null;
/* Base cases */
if (a == null)
return b;
if (b == null)
return a;
/* Pick either a or b, and recur */
if (a.val <= b.val) {
result = a;
result.next = sortedMerge(a.next, b);
} else {
result = b;
result.next = sortedMerge(a, b.next);
}
return result;
}
function mergeSort( h) {
// Base case : if head is null
if (h == null || h.next == null) {
return h;
}
// get the middle of the list
var middle = getMiddle(h);
var nextofmiddle = middle.next;
// set the next of middle node to null
middle.next = null;
// Apply mergeSort on left list
var left = mergeSort(h);
// Apply mergeSort on right list
var right = mergeSort(nextofmiddle);
// Merge the left and right lists
var sortedlist = sortedMerge(left, right);
return sortedlist;
}
// Utility function to get the middle
// of the linked list
function getMiddle( head) {
if (head == null)
return head;
var slow = head, fast = head;
while (fast.next != null && fast.next.next != null)
{
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
function push(new_data) {
/* allocate node */
var new_node = new node(new_data);
/* link the old list off the new node */
new_node.next = head;
/* move the head to point to the new node */
head = new_node;
}
// Utility function to print the linked list
function printList( headref) {
while (headref != null) {
document.write(headref.val + " ");
headref = headref.next;
}
}
/*
Let us create a unsorted linked
list to test the functions
created. The list shall be
a: 2->3->20->5->10->15
*/
push(15);
push(10);
push(5);
push(20);
push(3);
push(2);
// Apply merge Sort
head = mergeSort(head);
document.write("\n Sorted Linked List is: \n");
printList(head);
// This code contributed by umadevi9616
</script>
Producción:
Sorted Linked List is: 2 3 5 10 15 20
Complejidad de tiempo: O(n*log n)
Espacio Auxiliar: O(n)
Enfoque 2: este enfoque es más simple y utiliza el espacio log n.
mergeSort():
- Si el tamaño de la lista enlazada es 1, devuelve la cabeza
- Encuentre medio usando el enfoque de la Turtle y la liebre
- Almacene el siguiente de mid en head2, es decir, la lista de subenlaces correctos.
- Ahora haz que el siguiente punto medio sea nulo.
- Llame recursivamente a mergeSort() en la lista subvinculada izquierda y derecha y almacene el nuevo encabezado de la lista vinculada izquierda y derecha.
- Llame a merge() dados los argumentos nuevos encabezados de listas subvinculadas izquierda y derecha y almacene el encabezado final devuelto después de la fusión.
- Devuelve el encabezado final de la lista enlazada fusionada.
fusionar (cabeza1, cabeza2):
- Tome un puntero, digamos fusionado, para almacenar la lista fusionada en él y almacene un Node ficticio en él.
- Tome una temperatura de puntero y asígnele fusionar.
- Si los datos de head1 son menores que los datos de head2, entonces, almacene head1 en el siguiente de temp y mueva head1 al siguiente de head1.
- De lo contrario, almacene head2 en el siguiente de temp y mueva head2 al siguiente de head2.
- Mover temp al siguiente de temp.
- Repita los pasos 3, 4 y 5 hasta que head1 no sea igual a nulo y head2 no sea igual a nulo.
- Ahora agregue los Nodes restantes de la primera o la segunda lista vinculada a la lista vinculada fusionada.
- Devuelve el siguiente de fusionado (que ignorará el maniquí y devolverá el encabezado de la lista enlazada fusionada final)
C++
#include <iostream>
using namespace std;
// Node structure
struct Node {
int data;
Node* next;
};
// function to insert in list
void insert(int x, Node** head)
{
if (*head == NULL) {
*head = new Node;
(*head)->data = x;
(*head)->next = NULL;
return;
}
Node* temp = new Node;
temp->data = (*head)->data;
temp->next = (*head)->next;
(*head)->data = x;
(*head)->next = temp;
}
// function to print the list
void print(Node* head)
{
Node* temp = head;
while (temp != NULL) {
cout << temp->data << " ";
temp = temp->next;
}
}
Node* merge(Node* firstNode, Node* secondNode)
{
Node* merged = new Node;
Node* temp = new Node;
// merged is equal to temp so in the end we have the top
// Node.
merged = temp;
// while either firstNode or secondNode becomes NULL
while (firstNode != NULL && secondNode != NULL) {
if (firstNode->data <= secondNode->data) {
temp->next = firstNode;
firstNode = firstNode->next;
}
else {
temp->next = secondNode;
secondNode = secondNode->next;
}
temp = temp->next;
}
// any remaining Node in firstNode or secondNode gets
// inserted in the temp List
while (firstNode != NULL) {
temp->next = firstNode;
firstNode = firstNode->next;
temp = temp->next;
}
while (secondNode != NULL) {
temp->next = secondNode;
secondNode = secondNode->next;
temp = temp->next;
}
// return the head of the sorted list
return merged->next;
}
// function to calculate the middle Element
Node* middle(Node* head)
{
Node* slow = head;
Node* fast = head->next;
while (!slow->next && (!fast && !fast->next)) {
slow = slow->next;
fast = fast->next->next;
}
return slow;
}
// function to sort the given list
Node* sort(Node* head)
{
if (head->next == NULL) {
return head;
}
Node* mid = new Node;
Node* head2 = new Node;
mid = middle(head);
head2 = mid->next;
mid->next = NULL;
// recursive call to sort() hence diving our problem,
// and then merging the solution
Node* finalhead = merge(sort(head), sort(head2));
return finalhead;
}
int main(void)
{
Node* head = NULL;
int n[] = { 7, 10, 5, 20, 3, 2 };
for (int i = 0; i < 6; i++) {
insert(n[i], &head); // inserting in the list
}
cout << "Sorted Linked List is: \n";
print(sort(head)); // printing the sorted list returned
// by sort()
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
C
#include <stdio.h>
#include <stdlib.h>
// Node structure
typedef struct Node {
int data;
struct Node* next;
} Node;
// function to insert in list
void insert(int x, Node** head)
{
if (*head == NULL) {
*head = (Node*)malloc(sizeof(Node));
(*head)->data = x;
(*head)->next = NULL;
return;
}
Node* temp = (Node*)malloc(sizeof(Node));
temp->data = (*head)->data;
temp->next = (*head)->next;
(*head)->data = x;
(*head)->next = temp;
}
// function to print the list
void print(Node* head)
{
Node* temp = head;
while (temp != NULL) {
printf("%d ", temp->data);
temp = temp->next;
}
}
Node* merge(Node* firstNode, Node* secondNode)
{
Node* merged = (Node*)malloc(sizeof(Node));
Node* temp = (Node*)malloc(sizeof(Node));
// merged is equal to temp so in the end we have the top
// Node.
merged = temp;
// while either firstNode or secondNode becomes NULL
while (firstNode != NULL && secondNode != NULL) {
if (firstNode->data <= secondNode->data) {
temp->next = firstNode;
firstNode = firstNode->next;
}
else {
temp->next = secondNode;
secondNode = secondNode->next;
}
temp = temp->next;
}
// any remaining Node in firstNode or secondNode gets
// inserted in the temp List
while (firstNode != NULL) {
temp->next = firstNode;
firstNode = firstNode->next;
temp = temp->next;
}
while (secondNode != NULL) {
temp->next = secondNode;
secondNode = secondNode->next;
temp = temp->next;
}
// return the head of the sorted list
return merged->next;
}
// function to calculate the middle Element
Node* middle(Node* head)
{
Node* slow = head;
Node* fast = head->next;
while (!slow->next && (!fast && !fast->next)) {
slow = slow->next;
fast = fast->next->next;
}
return slow;
}
// function to sort the given list
Node* sort(Node* head)
{
if (head->next == NULL) {
return head;
}
Node* mid = (Node*)malloc(sizeof(Node));
Node* head2 = (Node*)malloc(sizeof(Node));
mid = middle(head);
head2 = mid->next;
mid->next = NULL;
// recursive call to sort() hence diving our problem,
// and then merging the solution
Node* finalhead = merge(sort(head), sort(head2));
return finalhead;
}
int main(void)
{
Node* head = NULL;
int n[] = { 7, 10, 5, 20, 3, 2 };
for (int i = 0; i < 6; i++) {
insert(n[i], &head); // inserting in the list
}
printf("Sorted Linked List is: \n");
print(sort(head)); // printing the sorted list returned
// by sort()
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
// Node Class
class Node {
int data;
Node next;
Node(int key)
{
this.data = key;
next = null;
}
}
class GFG {
// Function to merge sort
static Node mergeSort(Node head)
{
if (head.next == null)
return head;
Node mid = findMid(head);
Node head2 = mid.next;
mid.next = null;
Node newHead1 = mergeSort(head);
Node newHead2 = mergeSort(head2);
Node finalHead = merge(newHead1, newHead2);
return finalHead;
}
// Function to merge two linked lists
static Node merge(Node head1, Node head2)
{
Node merged = new Node(-1);
Node temp = merged;
// While head1 is not null and head2
// is not null
while (head1 != null && head2 != null) {
if (head1.data < head2.data) {
temp.next = head1;
head1 = head1.next;
}
else {
temp.next = head2;
head2 = head2.next;
}
temp = temp.next;
}
// While head1 is not null
while (head1 != null) {
temp.next = head1;
head1 = head1.next;
temp = temp.next;
}
// While head2 is not null
while (head2 != null) {
temp.next = head2;
head2 = head2.next;
temp = temp.next;
}
return merged.next;
}
// Find mid using The Tortoise and The Hare approach
static Node findMid(Node head)
{
Node slow = head, fast = head.next;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
// Function to print list
static void printList(Node head)
{
while (head != null) {
System.out.print(head.data + " ");
head = head.next;
}
}
// Driver Code
public static void main(String[] args)
{
Node head = new Node(7);
Node temp = head;
temp.next = new Node(10);
temp = temp.next;
temp.next = new Node(5);
temp = temp.next;
temp.next = new Node(20);
temp = temp.next;
temp.next = new Node(3);
temp = temp.next;
temp.next = new Node(2);
temp = temp.next;
// Apply merge Sort
head = mergeSort(head);
System.out.print("\nSorted Linked List is: \n");
printList(head);
}
}
Python3
# Python program for the above approach
# Node Class
class Node:
def __init__(self,key):
self.data=key
self.next=None
# Function to merge sort
def mergeSort(head):
if (head.next == None):
return head
mid = findMid(head)
head2 = mid.next
mid.next = None
newHead1 = mergeSort(head)
newHead2 = mergeSort(head2)
finalHead = merge(newHead1, newHead2)
return finalHead
# Function to merge two linked lists
def merge(head1,head2):
merged = Node(-1)
temp = merged
# While head1 is not null and head2
# is not null
while (head1 != None and head2 != None):
if (head1.data < head2.data):
temp.next = head1
head1 = head1.next
else:
temp.next = head2
head2 = head2.next
temp = temp.next
# While head1 is not null
while (head1 != None):
temp.next = head1
head1 = head1.next
temp = temp.next
# While head2 is not null
while (head2 != None):
temp.next = head2
head2 = head2.next
temp = temp.next
return merged.next
# Find mid using The Tortoise and The Hare approach
def findMid(head):
slow = head
fast = head.next
while (fast != None and fast.next != None):
slow = slow.next
fast = fast.next.next
return slow
# Function to print list
def printList(head):
while (head != None):
print(head.data,end=" ")
head=head.next
# Driver Code
head = Node(7)
temp = head
temp.next = Node(10);
temp = temp.next;
temp.next = Node(5);
temp = temp.next;
temp.next = Node(20);
temp = temp.next;
temp.next = Node(3);
temp = temp.next;
temp.next = Node(2);
temp = temp.next;
# Apply merge Sort
head = mergeSort(head);
print("\nSorted Linked List is: \n");
printList(head);
# This code is contributed by avanitrachhadiya2155
C#
// C# program for the above approach
using System;
// Node Class
public class Node
{
public int data;
public Node next;
public Node(int key)
{
this.data = key;
next = null;
}
}
class GFG{
// Function to merge sort
static Node mergeSort(Node head)
{
if (head.next == null)
return head;
Node mid = findMid(head);
Node head2 = mid.next;
mid.next = null;
Node newHead1 = mergeSort(head);
Node newHead2 = mergeSort(head2);
Node finalHead = merge(newHead1, newHead2);
return finalHead;
}
// Function to merge two linked lists
static Node merge(Node head1, Node head2)
{
Node merged = new Node(-1);
Node temp = merged;
// While head1 is not null and head2
// is not null
while (head1 != null && head2 != null)
{
if (head1.data < head2.data)
{
temp.next = head1;
head1 = head1.next;
}
else
{
temp.next = head2;
head2 = head2.next;
}
temp = temp.next;
}
// While head1 is not null
while (head1 != null)
{
temp.next = head1;
head1 = head1.next;
temp = temp.next;
}
// While head2 is not null
while (head2 != null)
{
temp.next = head2;
head2 = head2.next;
temp = temp.next;
}
return merged.next;
}
// Find mid using The Tortoise and The Hare approach
static Node findMid(Node head)
{
Node slow = head, fast = head.next;
while (fast != null && fast.next != null)
{
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
// Function to print list
static void printList(Node head)
{
while (head != null)
{
Console.Write(head.data + " ");
head = head.next;
}
}
// Driver Code
public static void Main(String[] args)
{
Node head = new Node(7);
Node temp = head;
temp.next = new Node(10);
temp = temp.next;
temp.next = new Node(5);
temp = temp.next;
temp.next = new Node(20);
temp = temp.next;
temp.next = new Node(3);
temp = temp.next;
temp.next = new Node(2);
temp = temp.next;
// Apply merge Sort
head = mergeSort(head);
Console.Write("\nSorted Linked List is: \n");
printList(head);
}
}
// This code is contributed by umadevi9616
Javascript
<script>
// JavaScript program for the above approach
// Node Class
class Node {
constructor(val) {
this.data = val;
this.next = null;
}
}
// Function to merge sort
function mergeSort(head) {
if (head.next == null)
return head;
var mid = findMid(head);
var head2 = mid.next;
mid.next = null;
var newHead1 = mergeSort(head);
var newHead2 = mergeSort(head2);
var finalHead = merge(newHead1, newHead2);
return finalHead;
}
// Function to merge two linked lists
function merge(head1, head2) {
var merged = new Node(-1);
var temp = merged;
// While head1 is not null and head2
// is not null
while (head1 != null && head2 != null) {
if (head1.data < head2.data) {
temp.next = head1;
head1 = head1.next;
} else {
temp.next = head2;
head2 = head2.next;
}
temp = temp.next;
}
// While head1 is not null
while (head1 != null) {
temp.next = head1;
head1 = head1.next;
temp = temp.next;
}
// While head2 is not null
while (head2 != null) {
temp.next = head2;
head2 = head2.next;
temp = temp.next;
}
return merged.next;
}
// Find mid using The Tortoise and The Hare approach
function findMid(head) {
var slow = head, fast = head.next;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
// Function to print list
function printList(head) {
while (head != null) {
document.write(head.data + " ");
head = head.next;
}
}
// Driver Code
var head = new Node(7);
var temp = head;
temp.next = new Node(10);
temp = temp.next;
temp.next = new Node(5);
temp = temp.next;
temp.next = new Node(20);
temp = temp.next;
temp.next = new Node(3);
temp = temp.next;
temp.next = new Node(2);
temp = temp.next;
// Apply merge Sort
head = mergeSort(head);
document.write("Sorted Linked List is: <br/>");
printList(head);
// This code contributed by gauravrajput1
</script>
Producción:
Sorted Linked List is: 2 3 5 7 10 20
Complejidad del tiempo : O(n*log n)
Espacio Auxiliar: O(n)
Fuentes:
http://en.wikipedia.org/wiki/Merge_sort
http://cslibrary.stanford.edu/105/LinkedListProblems.pdf
Escriba comentarios si encuentra que el código/algoritmo anterior es incorrecto, o encuentre mejores 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