合并两个未排序的链表得到一个排序的列表 – Set 2
给定两个未排序的链表， N个节点的L1和M个节点的L2 ，任务是将它们合并以得到一个排序的单链表。
例子：
Input: L1 = 3→5→1, L2 = 6→2→4→9
Output: 1→2→3→4→5→6→9
Input: L1 = 1→5→2, L2 = 3→7
Output: 1→2→3→5→7
注意：在 O(1) 辅助空间中解决此问题的内存高效方法已在此处接近，
方法：这个想法是使用一个辅助数组来存储两个链表的元素，然后按升序对数组进行排序。最后，将所有元素重新插入到链表中。请按照以下步骤解决问题：
- 通过将第一个列表的尾节点L1的下一个指向第二个列表的第一个节点L2来连接两个列表。
- 遍历链表L1 ，并将所有元素推入向量V 。
- 按升序对向量 V 进行排序。
- 排序后，将所有元素插入到列表L1中。
以下是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Linked list node
class Node {
public:
int data;
Node* next;
};
// Utility function to append key at
// end of linked list
void insertNode(Node** head, int x)
{
Node* ptr = new Node;
ptr->data = x;
ptr->next = NULL;
if (*head == NULL) {
*head = ptr;
}
else {
Node* temp;
temp = *head;
while (temp->next != NULL) {
temp = temp->next;
}
temp->next = ptr;
}
}
// Utility function to print linkedlist
void display(Node** head)
{
Node* temp;
temp = *head;
if (temp == NULL) {
cout << "NULL \n";
}
else {
while (temp != NULL) {
cout << temp->data;
if (temp->next != NULL)
cout << "->";
temp = temp->next;
}
}
}
// Function to merge two linked lists
void MergeLinkedlist(Node** head1, Node** head2)
{
Node* ptr;
ptr = *head1;
while (ptr->next != NULL) {
ptr = ptr->next;
}
// Join linked list by placing address of
// first node of L2 in the last node of L1
ptr->next = *head2;
}
// Function to merge two unsorted linked
// lists to get a sorted list
void sortLinkedList(Node** head, Node** head1)
{
// Function call to merge the two lists
MergeLinkedlist(head, head1);
// Declare a vector
vector V;
Node* ptr = *head;
// Push all elements into vector
while (ptr != NULL) {
V.push_back(ptr->data);
ptr = ptr->next;
}
// Sort the vector
sort(V.begin(), V.end());
int index = 0;
ptr = *head;
// Insert elements in the linked
// list from the vector
while (ptr != NULL) {
ptr->data = V[index];
index++;
ptr = ptr->next;
}
// Display the sorted and
// merged linked list
display(head);
}
// Driver Code
int main()
{
// Given linked list, L1
Node* head1 = NULL;
insertNode(&head1, 3);
insertNode(&head1, 5);
insertNode(&head1, 1);
// Given linked list, L2
Node* head2 = NULL;
insertNode(&head2, 6);
insertNode(&head2, 2);
insertNode(&head2, 4);
insertNode(&head2, 9);
// Function Call
sortLinkedList(&head1, &head2);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG {
// Linked list node
static class Node {
int data;
Node next;
};
static Node head1, head2;
// Utility function to append key at
// end of linked list
static Node insertNode(Node head, int x)
{
Node ptr = new Node();
ptr.data = x;
ptr.next = null;
if (head == null) {
head = ptr;
}
else {
Node temp;
temp = head;
while (temp.next != null) {
temp = temp.next;
}
temp.next = ptr;
}
return head;
}
// Utility function to print linkedlist
static void display(Node head)
{
Node temp;
temp = head;
if (temp == null) {
System.out.print("null \n");
}
else {
while (temp != null) {
System.out.print(temp.data);
if (temp.next != null)
System.out.print("->");
temp = temp.next;
}
}
}
// Function to merge two linked lists
static Node MergeLinkedlist()
{
Node ptr;
ptr = head1;
while (ptr.next != null) {
ptr = ptr.next;
}
// Join linked list by placing address of
// first node of L2 in the last node of L1
ptr.next = head2;
return head1;
}
// Function to merge two unsorted linked
// lists to get a sorted list
static void sortLinkedList()
{
// Function call to merge the two lists
Node head = MergeLinkedlist();
// Declare a vector
Vector V = new Vector<>();
Node ptr = head;
// Push all elements into vector
while (ptr != null) {
V.add(ptr.data);
ptr = ptr.next;
}
Collections.sort(V);
// Sort the vector
;
int index = 0;
ptr = head;
// Insert elements in the linked
// list from the vector
while (ptr != null) {
ptr.data = V.get(index);
index++;
ptr = ptr.next;
}
// Display the sorted and
// merged linked list
display(head);
}
// Driver Code
public static void main(String[] args)
{
// Given linked list, L1
head1 = insertNode(head1, 3);
head1 = insertNode(head1, 5);
head1 = insertNode(head1, 1);
// Given linked list, L2
head2 = null;
head2 = insertNode(head2, 6);
head2 = insertNode(head2, 2);
head2 = insertNode(head2, 4);
head2 = insertNode(head2, 9);
// Function Call
sortLinkedList();
}
}
// This code is contributed by umadevi9616 Python3
# Py program for the above approach
# Linked list node
class Node:
def __init__(self, d):
self.data = d
self.next = None
# Utility function to append key at
# end of linked list
def insertNode(head, x):
ptr = Node(x)
if (head == None):
head = ptr
else:
temp = head
while (temp.next != None):
temp = temp.next
temp.next = ptr
return head
# Utility function to print linkedlist
def display(head):
temp = head
if (temp == None):
print ("None")
else:
while (temp.next != None):
print (temp.data,end="->")
if (temp.next != None):
print ("",end="")
temp = temp.next
print(temp.data)
# Function to merge two linked lists
def MergeLinkedlist(head1, head2):
ptr = head1
while (ptr.next != None):
ptr = ptr.next
# Join linked list by placing address of
# first node of L2 in the last node of L1
ptr.next = head2
return head1
# Function to merge two unsorted linked
# lists to get a sorted list
def sortLinkedList(head1, head2):
# Function call to merge the two lists
head1 = MergeLinkedlist(head1, head2)
# Declare a vector
V = []
ptr = head1
# Push all elements into vector
while (ptr != None):
V.append(ptr.data)
ptr = ptr.next
# Sort the vector
V = sorted(V)
index = 0
ptr = head1
# Insert elements in the linked
# list from the vector
while (ptr != None):
ptr.data = V[index]
index += 1
ptr = ptr.next
# Display the sorted and
# merged linked list
display(head1)
# Driver Code
if __name__ == '__main__':
# Given linked list, L1
head1 = None
head1 = insertNode(head1, 3)
head1 = insertNode(head1, 5)
head1 = insertNode(head1, 1)
# Given linked list, L2
head2 = None
head2 = insertNode(head2, 6)
head2 = insertNode(head2, 2)
head2 = insertNode(head2, 4)
head2 = insertNode(head2, 9)
# Function Call
sortLinkedList(head1, head2)
# This code is contributed by mohit kumar 29.C#
// C# program for the above approach
using System;
using System.Collections.Generic;
public class GFG {
// Linked list node
public class Node {
public int data;
public Node next;
};
static Node head1, head2;
// Utility function to append key at
// end of linked list
static Node insertNode(Node head, int x) {
Node ptr = new Node();
ptr.data = x;
ptr.next = null;
if (head == null) {
head = ptr;
} else {
Node temp;
temp = head;
while (temp.next != null) {
temp = temp.next;
}
temp.next = ptr;
}
return head;
}
// Utility function to print linkedlist
static void display(Node head) {
Node temp;
temp = head;
if (temp == null) {
Console.Write("null \n");
} else {
while (temp != null) {
Console.Write(temp.data);
if (temp.next != null)
Console.Write("->");
temp = temp.next;
}
}
}
// Function to merge two linked lists
static Node MergeLinkedlist() {
Node ptr;
ptr = head1;
while (ptr.next != null) {
ptr = ptr.next;
}
// Join linked list by placing address of
// first node of L2 in the last node of L1
ptr.next = head2;
return head1;
}
// Function to merge two unsorted linked
// lists to get a sorted list
static void sortList()
{
// Function call to merge the two lists
Node head = MergeLinkedlist();
// Declare a vector
List V = new List();
Node ptr = head;
// Push all elements into vector
while (ptr != null) {
V.Add(ptr.data);
ptr = ptr.next;
}
V.Sort();
// Sort the vector
;
int index = 0;
ptr = head;
// Insert elements in the linked
// list from the vector
while (ptr != null) {
ptr.data = V[index];
index++;
ptr = ptr.next;
}
// Display the sorted and
// merged linked list
display(head);
}
// Driver Code
public static void Main(String[] args)
{
// Given linked list, L1
head1 = insertNode(head1, 3);
head1 = insertNode(head1, 5);
head1 = insertNode(head1, 1);
// Given linked list, L2
head2 = null;
head2 = insertNode(head2, 6);
head2 = insertNode(head2, 2);
head2 = insertNode(head2, 4);
head2 = insertNode(head2, 9);
// Function Call
sortList();
}
}
// This code is contributed by umadevi9616 Javascript
// Javascript program for the above approach
// Linked list node
class Node {
constructor(d) {
this.data = d
this.next = null
}
}
// Utility function to append key at
// end of linked list
function insertNode(head, x) {
ptr = new Node(x)
if (head == null) {
head = ptr
} else {
temp = head
while (temp.next != null) {
temp = temp.next
}
temp.next = ptr
}
return head
}
// Utility function to print linkedlist
function display(head) {
let temp = head
if (temp == null) {
document.write("null")
}
else {
while (temp.next != null) {
document.write(temp.data, end = "->")
if (temp.next != null)
document.write("", end = "")
temp = temp.next
}
document.write(temp.data)
}
}
// Function to merge two linked lists
function MergeLinkedlist(head1, head2) {
let ptr = head1
while (ptr.next != null)
ptr = ptr.next
// Join linked list by placing address of
// first node of L2 in the last node of L1
ptr.next = head2
return head1
}
// Function to merge two unsorted linked
// lists to get a sorted list
function sortLinkedList(head1, head2) {
// Function call to merge the two lists
head1 = MergeLinkedlist(head1, head2)
// Declare a vector
let V = []
let ptr = head1
// Push all elements into vector
while (ptr != null) {
V.push(ptr.data)
ptr = ptr.next
}
// Sort the vector
V = V.sort((a, b) => a - b)
index = 0
ptr = head1
// Insert elements in the linked
// list from the vector
while (ptr != null) {
ptr.data = V[index]
index += 1
ptr = ptr.next
}
// Display the sorted and
// merged linked list
display(head1)
}
// Driver Code
// Given linked list, L1
let head1 = null;
head1 = insertNode(head1, 3)
head1 = insertNode(head1, 5)
head1 = insertNode(head1, 1)
// Given linked list, L2
let head2 = null
head2 = insertNode(head2, 6)
head2 = insertNode(head2, 2)
head2 = insertNode(head2, 4)
head2 = insertNode(head2, 9)
// Function Call
sortLinkedList(head1, head2)
// This code is contributed by gfgking输出
1->2->3->4->5->6->9时间复杂度： O((N+M)*log(N+M))
辅助空间： O(N+M)