合并两个未排序的链接列表以获取一个已排序列表

给定两个未排序的链表，任务是将它们合并以获得一个排序的单链表。
例子：

Input: List 1 = 3 -> 1 -> 5, List 2 = 6-> 2 -> 4 
Output: 1 -> 2 -> 3 -> 4 -> 5 -> 6

Input: List 1 = 4 -> 7 -> 5, List 2 = 2-> 1 -> 8 -> 1 
Output: 1 -> 1 -> 2 -> 4 -> 5 -> 7 -> 8 
 

天真的方法：天真的方法是对给定的链表进行排序，然后将两个排序后的链表按升序合并到一个列表中。
为了解决上述问题，朴素的方法是对两个链表进行单独排序，并将两个链表合并为一个递增的列表。

高效方法：为了优化上述方法，我们将两个链表连接起来，然后使用任何排序算法对其进行排序。步骤如下：

1. 通过遍历第一个列表直到我们到达一个尾节点，然后将尾节点的下一个指向第二个列表的头节点，将两个列表连接起来。将此串联列表存储在第一个列表中。
2. 对以上合并的链表进行排序。在这里，我们将使用冒泡排序。因此，如果node-> next-> data小于node-> data，则交换两个相邻节点的数据。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Create structure for a node
struct node {
    int data;
    node* next;
};
 
// Function to print the linked list
void setData(node* head)
{
    node* tmp;
 
    // Store the head of the linked
    // list into a temporary node*
    // and iterate
    tmp = head;
 
    while (tmp != NULL) {
 
        cout << tmp->data
             << " -> ";
        tmp = tmp->next;
    }
}
 
// Function takes the head of the
// LinkedList and the data as
// argument and if no LinkedList
// exists, it creates one with the
// head pointing to first node.
// If it exists already, it appends
// given node at end of the last node
node* getData(node* head, int num)
{
 
    // Create a new node
    node* temp = new node;
    node* tail = head;
 
    // Insert data into the temporary
    // node and point it's next to NULL
    temp->data = num;
    temp->next = NULL;
 
    // Check if head is null, create a
    // linked list with temp as head
    // and tail of the list
    if (head == NULL) {
        head = temp;
        tail = temp;
    }
 
    // Else insert the temporary node
    // after the tail of the existing
    // node and make the temporary node
    // as the tail of the linked list
    else {
 
        while (tail != NULL) {
 
            if (tail->next == NULL) {
                tail->next = temp;
                tail = tail->next;
            }
            tail = tail->next;
        }
    }
 
    // Return the list
    return head;
}
 
// Function to concatenate the two lists
node* mergelists(node** head1,
                 node** head2)
{
 
    node* tail = *head1;
 
    // Iterate through the head1 to find the
    // last node join the next of last node
    // of head1 to the 1st node of head2
    while (tail != NULL) {
 
        if (tail->next == NULL
            && head2 != NULL) {
            tail->next = *head2;
            break;
        }
        tail = tail->next;
    }
 
    // return the concatenated lists as a
    // single list - head1
    return *head1;
}
 
// Sort the linked list using bubble sort
void sortlist(node** head1)
{
    node* curr = *head1;
    node* temp = *head1;
 
    // Compares two adjacent elements
    // and swaps if the first element
    // is greater than the other one.
    while (curr->next != NULL) {
 
        temp = curr->next;
        while (temp != NULL) {
 
            if (temp->data < curr->data) {
                int t = temp->data;
                temp->data = curr->data;
                curr->data = t;
            }
            temp = temp->next;
        }
        curr = curr->next;
    }
}
 
// Driver Code
int main()
{
    node* head1 = new node;
    node* head2 = new node;
 
    head1 = NULL;
    head2 = NULL;
 
    // Given Linked List 1
    head1 = getData(head1, 4);
    head1 = getData(head1, 7);
    head1 = getData(head1, 5);
 
    // Given Linked List 2
    head2 = getData(head2, 2);
    head2 = getData(head2, 1);
    head2 = getData(head2, 8);
    head2 = getData(head2, 1);
 
    // Merge the two lists
    // in a single list
    head1 = mergelists(&head1,
                       &head2);
 
    // Sort the unsorted merged list
    sortlist(&head1);
 
    // Print the final
    // sorted merged list
    setData(head1);
    return 0;
}

Java

// Java program for
// the above approach
class GFG{
 
static node head1 = null;
static node head2 = null;
   
// Create structure for a node
static class node
{
  int data;
  node next;
};
 
// Function to print
// the linked list
static void setData(node head)
{
  node tmp;
   
  // Store the head of the linked
  // list into a temporary node
  // and iterate
  tmp = head;
 
  while (tmp != null)
  {
    System.out.print(tmp.data + " -> ");
    tmp = tmp.next;
  }
}
 
// Function takes the head of the
// LinkedList and the data as
// argument and if no LinkedList
// exists, it creates one with the
// head pointing to first node.
// If it exists already, it appends
// given node at end of the last node
static node getData(node head, int num)
{
  // Create a new node
  node temp = new node();
  node tail = head;
 
  // Insert data into the temporary
  // node and point it's next to null
  temp.data = num;
  temp.next = null;
 
  // Check if head is null, create a
  // linked list with temp as head
  // and tail of the list
  if (head == null)
  {
    head = temp;
    tail = temp;
  }
 
  // Else insert the temporary node
  // after the tail of the existing
  // node and make the temporary node
  // as the tail of the linked list
  else
  {
    while (tail != null)
    {
      if (tail.next == null)
      {
        tail.next = temp;
        tail = tail.next;
      }
      tail = tail.next;
    }
  }
 
  // Return the list
  return head;
}
 
// Function to concatenate
// the two lists
static node mergelists()
{
  node tail = head1;
 
  // Iterate through the
  // head1 to find the
  // last node join the
  // next of last node
  // of head1 to the
  // 1st node of head2
  while (tail != null)
  {
    if (tail.next == null &&
        head2 != null)
    {
      tail.next = head2;
      break;
    }
    tail = tail.next;
  }
 
  // return the concatenated
  // lists as a single list - head1
  return head1;
}
 
// Sort the linked list
// using bubble sort
static void sortlist()
{
  node curr = head1;
  node temp = head1;
 
  // Compares two adjacent elements
  // and swaps if the first element
  // is greater than the other one.
  while (curr.next != null)
  {
    temp = curr.next;
    while (temp != null)
    {
      if (temp.data < curr.data)
      {
        int t = temp.data;
        temp.data = curr.data;
        curr.data = t;
      }
      temp = temp.next;
    }
    curr = curr.next;
  }
}
 
// Driver Code
public static void main(String[] args)
{
  // Given Linked List 1
  head1 = getData(head1, 4);
  head1 = getData(head1, 7);
  head1 = getData(head1, 5);
 
  // Given Linked List 2
  head2 = getData(head2, 2);
  head2 = getData(head2, 1);
  head2 = getData(head2, 8);
  head2 = getData(head2, 1);
 
  // Merge the two lists
  // in a single list
  head1 = mergelists();
 
  // Sort the unsorted merged list
  sortlist();
 
  // Print the final
  // sorted merged list
  setData(head1);
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python3 program for the
# above approach
 
# Create structure for a node
class node:
   
    def __init__(self, x):
       
        self.data = x
        self.next = None
 
# Function to print the linked
# list
def setData(head):
 
    # Store the head of the
    # linked list into a
    # temporary node* and
    # iterate
    tmp = head
 
    while (tmp != None):
        print(tmp.data,
              end = " -> ")
        tmp = tmp.next
 
# Function takes the head of the
# LinkedList and the data as
# argument and if no LinkedList
# exists, it creates one with the
# head pointing to first node.
# If it exists already, it appends
# given node at end of the last node
def getData(head, num):
 
    # Create a new node
    temp = node(-1)
    tail = head
 
    # Insert data into the temporary
    # node and point it's next to NULL
    temp.data = num
    temp.next = None
 
    # Check if head is null, create a
    # linked list with temp as head
    # and tail of the list
    if (head == None):
        head = temp
        tail = temp
 
    # Else insert the temporary node
    # after the tail of the existing
    # node and make the temporary node
    # as the tail of the linked list
    else:
        while (tail != None):
            if (tail.next == None):
                tail.next = temp
                tail = tail.next
            tail = tail.next
 
    # Return the list
    return head
 
# Function to concatenate the
# two lists
def mergelists(head1,head2):
 
    tail = head1
 
    # Iterate through the head1 to
    # find the last node join the
    # next of last node of head1
    # to the 1st node of head2
    while (tail != None):
        if (tail.next == None
            and head2 != None):
            tail.next =head2
            break
        tail = tail.next
 
    # return the concatenated
    # lists as a single list
    # - head1
    return head1
 
# Sort the linked list using
# bubble sort
def sortlist(head1):
   
    curr = head1
    temp = head1
 
    # Compares two adjacent elements
    # and swaps if the first element
    # is greater than the other one.
    while (curr.next != None):
        temp = curr.next
        while (temp != None):
            if (temp.data < curr.data):
                t = temp.data
                temp.data = curr.data
                curr.data = t
            temp = temp.next
        curr = curr.next
 
# Driver Code
if __name__ == '__main__':
   
    head1 = node(-1)
    head2 = node(-1)
 
    head1 = None
    head2 = None
 
    # Given Linked List 1
    head1 = getData(head1, 4)
    head1 = getData(head1, 7)
    head1 = getData(head1, 5)
 
    # Given Linked List 2
    head2 = getData(head2, 2)
    head2 = getData(head2, 1)
    head2 = getData(head2, 8)
    head2 = getData(head2, 1)
 
    # Merge the two lists
    # in a single list
    head1 = mergelists(head1,head2)
 
    # Sort the unsorted merged list
    sortlist(head1)
 
    # Print the final
    # sorted merged list
    setData(head1)
 
# This code is contributed by Mohit Kumar 29

C#

// C# program for
// the above approach
using System;
 
class GFG{
 
static node head1 = null;
static node head2 = null;
   
// Create structure for a node
class node
{
    public int data;
    public node next;
};
 
// Function to print
// the linked list
static void setData(node head)
{
    node tmp;
     
    // Store the head of the linked
    // list into a temporary node
    // and iterate
    tmp = head;
     
    while (tmp != null)
    {
        Console.Write(tmp.data + " -> ");
        tmp = tmp.next;
    }
}
 
// Function takes the head of
//the List and the data as
// argument and if no List
// exists, it creates one with the
// head pointing to first node.
// If it exists already, it appends
// given node at end of the last node
static node getData(node head, int num)
{
     
    // Create a new node
    node temp = new node();
    node tail = head;
     
    // Insert data into the temporary
    // node and point it's next to null
    temp.data = num;
    temp.next = null;
     
    // Check if head is null, create a
    // linked list with temp as head
    // and tail of the list
    if (head == null)
    {
        head = temp;
        tail = temp;
    }
     
    // Else insert the temporary node
    // after the tail of the existing
    // node and make the temporary node
    // as the tail of the linked list
    else
    {
        while (tail != null)
        {
            if (tail.next == null)
            {
                tail.next = temp;
                tail = tail.next;
            }
            tail = tail.next;
        }
    }
     
    // Return the list
    return head;
}
 
// Function to concatenate
// the two lists
static node mergelists()
{
    node tail = head1;
     
    // Iterate through the
    // head1 to find the
    // last node join the
    // next of last node
    // of head1 to the
    // 1st node of head2
    while (tail != null)
    {
        if (tail.next == null &&
                head2 != null)
        {
            tail.next = head2;
            break;
        }
        tail = tail.next;
    }
     
    // return the concatenated
    // lists as a single list - head1
    return head1;
}
 
// Sort the linked list
// using bubble sort
static void sortlist()
{
    node curr = head1;
    node temp = head1;
     
    // Compares two adjacent elements
    // and swaps if the first element
    // is greater than the other one.
    while (curr.next != null)
    {
        temp = curr.next;
        while (temp != null)
        {
            if (temp.data < curr.data)
            {
                int t = temp.data;
                temp.data = curr.data;
                curr.data = t;
            }
            temp = temp.next;
        }
        curr = curr.next;
    }
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given Linked List 1
    head1 = getData(head1, 4);
    head1 = getData(head1, 7);
    head1 = getData(head1, 5);
     
    // Given Linked List 2
    head2 = getData(head2, 2);
    head2 = getData(head2, 1);
    head2 = getData(head2, 8);
    head2 = getData(head2, 1);
     
    // Merge the two lists
    // in a single list
    head1 = mergelists();
     
    // Sort the unsorted merged list
    sortlist();
     
    // Print the final
    // sorted merged list
    setData(head1);
}
}
 
// This code is contributed by Amit Katiyar

输出：

1 -> 1 -> 2 -> 4 -> 5 -> 7 -> 8

时间复杂度： O(M * N) ，其中M和N是两个给定链表的长度。