合并 K 排序链表 |设置 1

给定 K 个大小为 N 的排序链表，合并它们并打印排序后的输出。

例子：

Input: k = 3, n =  4
list1 = 1->3->5->7->NULL
list2 = 2->4->6->8->NULL
list3 = 0->9->10->11->NULL

Output: 0->1->2->3->4->5->6->7->8->9->10->11
Merged lists in a sorted order 
where every element is greater 
than the previous element.

Input: k = 3, n =  3
list1 = 1->3->7->NULL
list2 = 2->4->8->NULL
list3 = 9->10->11->NULL

Output: 1->2->3->4->7->8->9->10->11
Merged lists in a sorted order 
where every element is greater 
than the previous element.

方法一（简单）

方法：
一个简单的解决方案是将结果初始化为第一个列表。现在从第二个列表开始遍历所有列表。以排序的方式将当前遍历的列表的每个节点插入到结果中。

C++

// C++ program to merge k sorted
// arrays of size n each
#include 
using namespace std;
 
// A Linked List node
struct Node {
    int data;
    Node* next;
};
 
/* Function to print nodes in
   a given linked list */
void printList(Node* node)
{
    while (node != NULL) {
        printf("%d ", node->data);
        node = node->next;
    }
}
 
// The main function that
// takes an array of lists
// arr[0..last] and generates
// the sorted output
Node* mergeKLists(Node* arr[], int last)
{
 
    // Traverse form second list to last
    for (int i = 1; i <= last; i++) {
        while (true) {
            // head of both the lists,
            // 0 and ith list.
            Node *head_0 = arr[0], *head_i = arr[i];
 
            // Break if list ended
            if (head_i == NULL)
                break;
 
            // Smaller than first element
            if (head_0->data >= head_i->data) {
                arr[i] = head_i->next;
                head_i->next = head_0;
                arr[0] = head_i;
            }
            else
                // Traverse the first list
                while (head_0->next != NULL) {
                    // Smaller than next element
                    if (head_0->next->data
                        >= head_i->data) {
                        arr[i] = head_i->next;
                        head_i->next = head_0->next;
                        head_0->next = head_i;
                        break;
                    }
                    // go to next node
                    head_0 = head_0->next;
 
                    // if last node
                    if (head_0->next == NULL) {
                        arr[i] = head_i->next;
                        head_i->next = NULL;
                        head_0->next = head_i;
                        head_0->next->next = NULL;
                        break;
                    }
                }
        }
    }
 
    return arr[0];
}
 
// Utility function to create a new node.
Node* newNode(int data)
{
    struct Node* temp = new Node;
    temp->data = data;
    temp->next = NULL;
    return temp;
}
 
// Driver program to test
// above functions
int main()
{
    // Number of linked lists
    int k = 3;
 
    // Number of elements in each list
    int n = 4;
 
    // an array of pointers storing the
    // head nodes of the linked lists
    Node* arr[k];
 
    arr[0] = newNode(1);
    arr[0]->next = newNode(3);
    arr[0]->next->next = newNode(5);
    arr[0]->next->next->next = newNode(7);
 
    arr[1] = newNode(2);
    arr[1]->next = newNode(4);
    arr[1]->next->next = newNode(6);
    arr[1]->next->next->next = newNode(8);
 
    arr[2] = newNode(0);
    arr[2]->next = newNode(9);
    arr[2]->next->next = newNode(10);
    arr[2]->next->next->next = newNode(11);
 
    // Merge all lists
    Node* head = mergeKLists(arr, k - 1);
 
    printList(head);
 
    return 0;
}

Java

// Java program to merge k sorted
// arrays of size n each
import java.io.*;
 
// A Linked List node
class Node
{
  int data;
  Node next;
 
  // Utility function to create a new node.
  Node(int key)
  {
    data = key;
    next = null;
  }
}
class GFG {
 
  static Node head;
  static Node temp;
 
  /* Function to print nodes in
   a given linked list */
  static void printList(Node node)
  {
    while(node != null)
    {
      System.out.print(node.data + " ");
 
      node = node.next;
    }
    System.out.println();
  }
 
  // The main function that
  // takes an array of lists
  // arr[0..last] and generates
  // the sorted output
  static Node mergeKLists(Node arr[], int last)
  {
 
    // Traverse form second list to last
    for (int i = 1; i <= last; i++)
    {
      while(true)
      {
 
        // head of both the lists,
        // 0 and ith list. 
        Node head_0 = arr[0];
        Node head_i = arr[i];
 
        // Break if list ended
        if (head_i == null)
          break;
 
        // Smaller than first element
        if(head_0.data >= head_i.data)
        {
          arr[i] = head_i.next;
          head_i.next = head_0;
          arr[0] = head_i;
        }
        else
        {
 
          // Traverse the first list
          while (head_0.next != null)
          {
 
            // Smaller than next element
            if (head_0.next.data >= head_i.data)
            {
              arr[i] = head_i.next;
              head_i.next = head_0.next;
              head_0.next = head_i;
              break;
            }
 
            // go to next node
            head_0 = head_0.next;
 
            // if last node
            if (head_0.next == null)
            {
              arr[i] = head_i.next;
              head_i.next = null;
              head_0.next = head_i;
              head_0.next.next = null;
              break;
            }
          }
        }
      }
    }
    return arr[0];
  }
 
  // Driver program to test
  // above functions 
  public static void main (String[] args)
  {
 
    // Number of linked lists
    int k = 3;
 
    // Number of elements in each list
    int n = 4;
 
    // an array of pointers storing the
    // head nodes of the linked lists
 
    Node[] arr = new Node[k];
 
    arr[0] = new Node(1);
    arr[0].next = new Node(3);
    arr[0].next.next = new Node(5);
    arr[0].next.next.next = new Node(7);
 
    arr[1] = new Node(2);
    arr[1].next = new Node(4);
    arr[1].next.next = new Node(6);
    arr[1].next.next.next = new Node(8);
 
    arr[2] = new Node(0);
    arr[2].next = new Node(9);
    arr[2].next.next = new Node(10);
    arr[2].next.next.next = new Node(11);
 
    // Merge all lists
    head = mergeKLists(arr, k - 1);
    printList(head);
 
  }
}
 
// This code is contributed by avanitrachhadiya2155

Python3

# Python3 program to merge k
# sorted arrays of size n each
 
# A Linked List node
class Node:
   
    def __init__(self, x):
       
        self.data = x
        self.next = None
 
# Function to prnodes in
# a given linked list
def printList(node):
   
    while (node != None):
        print(node.data,
              end = " ")
        node = node.next
 
# The main function that
# takes an array of lists
# arr[0..last] and generates
# the sorted output
def mergeKLists(arr, last):
 
    # Traverse form second
    # list to last
    for i in range(1, last + 1):
        while (True):
            # head of both the lists,
            # 0 and ith list.
            head_0 = arr[0]
            head_i = arr[i]
 
            # Break if list ended
            if (head_i == None):
                break
 
            # Smaller than first
            # element
            if (head_0.data >=
                head_i.data):
                arr[i] = head_i.next
                head_i.next = head_0
                arr[0] = head_i
            else:
                # Traverse the first list
                while (head_0.next != None):
                    # Smaller than next
                    # element
                    if (head_0.next.data >=
                        head_i.data):
                        arr[i] = head_i.next
                        head_i.next = head_0.next
                        head_0.next = head_i
                        break
                    # go to next node
                    head_0 = head_0.next
                    # if last node
                    if (head_0.next == None):
                        arr[i] = head_i.next
                        head_i.next = None
                        head_0.next = head_i
                        head_0.next.next = None
                        break
    return arr[0]
 
# Driver code
if __name__ == '__main__':
   
    # Number of linked
    # lists
    k = 3
     
    # Number of elements
    # in each list
    n = 4
 
    # an array of pointers
    # storing the head nodes
    # of the linked lists
    arr = [None for i in range(k)]
 
    arr[0] = Node(1)
    arr[0].next = Node(3)
    arr[0].next.next = Node(5)
    arr[0].next.next.next = Node(7)
 
    arr[1] = Node(2)
    arr[1].next = Node(4)
    arr[1].next.next = Node(6)
    arr[1].next.next.next = Node(8)
 
    arr[2] = Node(0)
    arr[2].next = Node(9)
    arr[2].next.next = Node(10)
    arr[2].next.next.next = Node(11)
 
    # Merge all lists
    head = mergeKLists(arr, k - 1)
 
    printList(head)
 
# This code is contributed by Mohit Kumar 29

C#

// C# program to merge k sorted
// arrays of size n each
using System;
 
// A Linked List node
public class Node
{
  public int data;
  public Node next;
 
  // Utility function to create a new node.
  public Node(int key)
  {
    data = key;
    next = null;
  }
}
 
public class GFG
{
  static Node head;
 
  /* Function to print nodes in
   a given linked list */
  static void printList(Node node)
  {
    while(node != null)
    {
      Console.Write(node.data + " ");
      node = node.next;
    }
    Console.WriteLine();
  }
 
  // The main function that
  // takes an array of lists
  // arr[0..last] and generates
  // the sorted output
  static Node mergeKLists(Node[] arr, int last)
  {
 
    // Traverse form second list to last
    for (int i = 1; i <= last; i++)
    {
      while(true)
      {
 
        // head of both the lists,
        // 0 and ith list. 
        Node head_0 = arr[0];
        Node head_i = arr[i];
 
        // Break if list ended
        if (head_i == null)
          break;
 
        // Smaller than first element
        if(head_0.data >= head_i.data)
        {
          arr[i] = head_i.next;
          head_i.next = head_0;
          arr[0] = head_i;
        }
        else
        {
 
          // Traverse the first list
          while (head_0.next != null)
          {
 
            // Smaller than next element
            if (head_0.next.data >= head_i.data)
            {
              arr[i] = head_i.next;
              head_i.next = head_0.next;
              head_0.next = head_i;
              break;
            }
 
            // go to next node
            head_0 = head_0.next;
 
            // if last node
            if (head_0.next == null)
            {
              arr[i] = head_i.next;
              head_i.next = null;
              head_0.next = head_i;
              head_0.next.next = null;
              break;
            }
          }
        }
      }
    }
    return arr[0];
  }
  static public void Main ()
  {
 
    // Number of linked lists
    int k = 3;
 
    // an array of pointers storing the
    // head nodes of the linked lists
    Node[] arr = new Node[k];
 
    arr[0] = new Node(1);
    arr[0].next = new Node(3);
    arr[0].next.next = new Node(5);
    arr[0].next.next.next = new Node(7);
 
    arr[1] = new Node(2);
    arr[1].next = new Node(4);
    arr[1].next.next = new Node(6);
    arr[1].next.next.next = new Node(8);
 
    arr[2] = new Node(0);
    arr[2].next = new Node(9);
    arr[2].next.next = new Node(10);
    arr[2].next.next.next = new Node(11);
 
    // Merge all lists
    head = mergeKLists(arr, k - 1);
    printList(head);
  }
}
 
// This code is contributed by rag2127

Javascript

// Javascript program to merge k sorted
// arrays of size n each
     
    // A Linked List node
    class Node
    {
        // Utility function to create a new node.
        constructor(key)
        {
            this.data=key;
            this.next=null;
        }
    }
     
    let head;
    let temp;
 
/* Function to print nodes in
a given linked list */
function printList(node)
{
    while(node != null)
    {
      document.write(node.data + " ");
  
      node = node.next;
    }
    document.write("<br>");
         
}
 
// The main function that
// takes an array of lists
// arr[0..last] and generates
// the sorted output
function mergeKLists(arr,last)
{
    // Traverse form second list to last
    for (let i = 1; i <= last; i++)
    {
      while(true)
      {
  
        // head of both the lists,
        // 0 and ith list.
        let head_0 = arr[0];
        let head_i = arr[i];
  
        // Break if list ended
        if (head_i == null)
          break;
  
        // Smaller than first element
        if(head_0.data >= head_i.data)
        {
          arr[i] = head_i.next;
          head_i.next = head_0;
          arr[0] = head_i;
        }
        else
        {
  
          // Traverse the first list
          while (head_0.next != null)
          {
  
            // Smaller than next element
            if (head_0.next.data >= head_i.data)
            {
              arr[i] = head_i.next;
              head_i.next = head_0.next;
              head_0.next = head_i;
              break;
            }
  
            // go to next node
            head_0 = head_0.next;
  
            // if last node
            if (head_0.next == null)
            {
              arr[i] = head_i.next;
              head_i.next = null;
              head_0.next = head_i;
              head_0.next.next = null;
              break;
            }
          }
        }
      }
    }
    return arr[0];
}
 
// Driver program to test
// above functions
// Number of linked lists
let k = 3;
// Number of elements in each list
let n = 4;
// an array of pointers storing the
// head nodes of the linked lists
let  arr = new Array(k);
arr[0] = new Node(1);
arr[0].next = new Node(3);
arr[0].next.next = new Node(5);
arr[0].next.next.next = new Node(7);
 
arr[1] = new Node(2);
arr[1].next = new Node(4);
arr[1].next.next = new Node(6);
arr[1].next.next.next = new Node(8);
 
arr[2] = new Node(0);
arr[2].next = new Node(9);
arr[2].next.next = new Node(10);
arr[2].next.next.next = new Node(11);
 
// Merge all lists
head = mergeKLists(arr, k - 1);
printList(head);
 
     
 
// This code is contributed by unknown2108
 

输出：

0 1 2 3 4 5 6 7 8 9 10 11

复杂度分析：

• 时间复杂度： O(nk ² )
• 辅助空间： O(1)。
  因为不需要额外的空间。

方法二：最小堆。
更好的解决方案是使用基于最小堆的解决方案，此处针对数组进行了讨论。该解决方案的时间复杂度为O(nk Log k)方法三：分而治之。
在这篇文章中，讨论了分而治之的方法。这种方法不需要额外的堆空间，并且在 O(nk Log k) 中工作
众所周知，两个链表的合并可以在 O(n) 时间和 O(n) 空间内完成。

1. 这个想法是配对 K 个列表并使用 O(n) 空间在线性时间内合并每一对。
2. 在第一个循环之后，剩下 K/2 个列表，每个列表的大小为 2*N。在第二个循环之后，剩下 K/4 个列表，每个列表的大小为 4*N，依此类推。
3. 重复这个过程，直到我们只剩下一个列表。

下面是上述想法的实现。

C++

// C++ program to merge k sorted
// arrays of size n each
#include 
using namespace std;
 
// A Linked List node
struct Node {
    int data;
    Node* next;
};
 
/* Function to print nodes in
   a given linked list */
void printList(Node* node)
{
    while (node != NULL) {
        printf("%d ", node->data);
        node = node->next;
    }
}
 
/* Takes two lists sorted in increasing order, and merge
   their nodes together to make one big sorted list. Below
   function takes O(n) extra space for recursive calls,
    */
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;
}
 
// The main function that takes an array of lists
// arr[0..last] and generates the sorted output
Node* mergeKLists(Node* arr[], int last)
{
    // repeat until only one list is left
    while (last != 0) {
        int i = 0, j = last;
 
        // (i, j) forms a pair
        while (i < j) {
            // merge List i with List j and store
            // merged list in List i
            arr[i] = SortedMerge(arr[i], arr[j]);
 
            // consider next pair
            i++, j--;
 
            // If all pairs are merged, update last
            if (i >= j)
                last = j;
        }
    }
 
    return arr[0];
}
 
// Utility function to create a new node.
Node* newNode(int data)
{
    struct Node* temp = new Node;
    temp->data = data;
    temp->next = NULL;
    return temp;
}
 
// Driver program to test above functions
int main()
{
    int k = 3; // Number of linked lists
    int n = 4; // Number of elements in each list
 
    // an array of pointers storing the head nodes
    // of the linked lists
    Node* arr[k];
 
    arr[0] = newNode(1);
    arr[0]->next = newNode(3);
    arr[0]->next->next = newNode(5);
    arr[0]->next->next->next = newNode(7);
 
    arr[1] = newNode(2);
    arr[1]->next = newNode(4);
    arr[1]->next->next = newNode(6);
    arr[1]->next->next->next = newNode(8);
 
    arr[2] = newNode(0);
    arr[2]->next = newNode(9);
    arr[2]->next->next = newNode(10);
    arr[2]->next->next->next = newNode(11);
 
    // Merge all lists
    Node* head = mergeKLists(arr, k - 1);
 
    printList(head);
 
    return 0;
}

Java

// Java program to merge k sorted arrays of size n each
public class MergeKSortedLists {
 
    /* Takes two lists sorted in increasing order, and merge
    their nodes together to make one big sorted list. Below
    function takes O(Log n) extra space for recursive calls,
    but it can be easily modified to work with same time and
    O(1) extra space  */
    public static 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;
    }
 
    // The main function that takes an array of lists
    // arr[0..last] and generates the sorted output
    public static Node mergeKLists(Node arr[], int last)
    {
        // repeat until only one list is left
        while (last != 0) {
            int i = 0, j = last;
 
            // (i, j) forms a pair
            while (i < j) {
                // merge List i with List j and store
                // merged list in List i
                arr[i] = SortedMerge(arr[i], arr[j]);
 
                // consider next pair
                i++;
                j--;
 
                // If all pairs are merged, update last
                if (i >= j)
                    last = j;
            }
        }
 
        return arr[0];
    }
 
    /* Function to print nodes in a given linked list */
    public static void printList(Node node)
    {
        while (node != null) {
            System.out.print(node.data + " ");
            node = node.next;
        }
    }
 
    public static void main(String args[])
    {
        int k = 3; // Number of linked lists
        int n = 4; // Number of elements in each list
 
        // an array of pointers storing the head nodes
        // of the linked lists
        Node arr[] = new Node[k];
 
        arr[0] = new Node(1);
        arr[0].next = new Node(3);
        arr[0].next.next = new Node(5);
        arr[0].next.next.next = new Node(7);
 
        arr[1] = new Node(2);
        arr[1].next = new Node(4);
        arr[1].next.next = new Node(6);
        arr[1].next.next.next = new Node(8);
 
        arr[2] = new Node(0);
        arr[2].next = new Node(9);
        arr[2].next.next = new Node(10);
        arr[2].next.next.next = new Node(11);
 
        // Merge all lists
        Node head = mergeKLists(arr, k - 1);
        printList(head);
    }
}
 
class Node {
    int data;
    Node next;
    Node(int data)
    {
        this.data = data;
    }
}
// This code is contributed by Gaurav Tiwari

蟒蛇3

# Python3 program to merge k sorted
# arrays of size n each
  
# A Linked List node
class Node:
     
    def __init__(self):
         
        self.data = 0
        self.next = None
 
# Function to print nodes in a
# given linked list
def printList(node):
 
    while (node != None):
        print(node.data, end = ' ')
        node = node.next
     
# Takes two lists sorted in increasing order,
# and merge their nodes together to make one
# big sorted list. Below function takes
# O(Log n) extra space for recursive calls,
# but it can be easily modified to work with
# same time and O(1) extra space
def SortedMerge(a, b):
 
    result = None
  
    # Base cases
    if (a == None):
        return(b)
    elif (b == None):
        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
 
# The main function that takes an array of lists
# arr[0..last] and generates the sorted output
def mergeKLists(arr, last):
 
    # Repeat until only one list is left
    while (last != 0):
        i = 0
        j = last
  
        # (i, j) forms a pair
        while (i < j):
             
            # Merge List i with List j and store
            # merged list in List i
            arr[i] = SortedMerge(arr[i], arr[j])
  
            # Consider next pair
            i += 1
            j -= 1
             
            # If all pairs are merged, update last
            if (i >= j):
                last = j
  
    return arr[0]
 
# Utility function to create a new node.
def newNode(data):
 
    temp = Node()
    temp.data = data
    temp.next = None
    return temp
 
# Driver code
if __name__=='__main__':
     
    # Number of linked lists
    k = 3
     
    # Number of elements in each list
    n = 4
  
    # An array of pointers storing the
    # head nodes of the linked lists
    arr = [0 for i in range(k)]
  
    arr[0] = newNode(1)
    arr[0].next = newNode(3)
    arr[0].next.next = newNode(5)
    arr[0].next.next.next = newNode(7)
  
    arr[1] = newNode(2)
    arr[1].next = newNode(4)
    arr[1].next.next = newNode(6)
    arr[1].next.next.next = newNode(8)
  
    arr[2] = newNode(0)
    arr[2].next = newNode(9)
    arr[2].next.next = newNode(10)
    arr[2].next.next.next = newNode(11)
  
    # Merge all lists
    head = mergeKLists(arr, k - 1)
  
    printList(head)
 
# This code is contributed by rutvik_56

C＃

// C# program to merge k sorted arrays of size n each
using System;
 
public class MergeKSortedLists {
 
    /* Takes two lists sorted in
    increasing order, and merge
    their nodes together to make
    one big sorted list. Below
    function takes O(Log n) extra
    space for recursive calls,
    but it can be easily modified
    to work with same time and
    O(1) extra space */
    public static 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;
    }
 
    // The main function that takes
    // an array of lists arr[0..last]
    // and generates the sorted output
    public static Node mergeKLists(Node[] arr, int last)
    {
        // repeat until only one list is left
        while (last != 0) {
            int i = 0, j = last;
 
            // (i, j) forms a pair
            while (i < j) {
                // merge List i with List j and store
                // merged list in List i
                arr[i] = SortedMerge(arr[i], arr[j]);
 
                // consider next pair
                i++;
                j--;
 
                // If all pairs are merged, update last
                if (i >= j)
                    last = j;
            }
        }
 
        return arr[0];
    }
 
    /* Function to print nodes in a given linked list */
    public static void printList(Node node)
    {
        while (node != null) {
            Console.Write(node.data + " ");
            node = node.next;
        }
    }
 
    public static void Main()
    {
        int k = 3; // Number of linked lists
        //int n = 4; // Number of elements in each list
 
        // An array of pointers storing the head nodes
        // of the linked lists
        Node[] arr = new Node[k];
 
        arr[0] = new Node(1);
        arr[0].next = new Node(3);
        arr[0].next.next = new Node(5);
        arr[0].next.next.next = new Node(7);
 
        arr[1] = new Node(2);
        arr[1].next = new Node(4);
        arr[1].next.next = new Node(6);
        arr[1].next.next.next = new Node(8);
 
        arr[2] = new Node(0);
        arr[2].next = new Node(9);
        arr[2].next.next = new Node(10);
        arr[2].next.next.next = new Node(11);
 
        // Merge all lists
        Node head = mergeKLists(arr, k - 1);
        printList(head);
    }
}
 
public class Node {
    public int data;
    public Node next;
    public Node(int data)
    {
        this.data = data;
    }
}
 
/* This code contributed by PrinciRaj1992 */

Javascript

 
// JavaScript program to merge k sorted
// arrays of size n each
class Node {
    constructor(val) {
        this.data = val;
        this.next = null;
    }
}
 
    /*
      Takes two lists sorted in increasing order,
     and merge their nodes together to
     make one big sorted list.
     Below function takes O(Log n) extra space for
     * recursive calls, but it can be easily modified
     to work with same time and
     * O(1) extra space
     */
    function SortedMerge(a,  b) {
var 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;
    }
 
    // The main function that takes an array of lists
    // arr[0..last] and generates the sorted output
    function mergeKLists(arr , last) {
        // repeat until only one list is left
        while (last != 0) {
            var i = 0, j = last;
 
            // (i, j) forms a pair
            while (i < j) {
                // merge List i with List j and store
                // merged list in List i
                arr[i] = SortedMerge(arr[i], arr[j]);
 
                // consider next pair
                i++;
                j--;
 
                // If all pairs are merged, update last
                if (i >= j)
                    last = j;
            }
        }
 
        return arr[0];
    }
 
    /* Function to prvar nodes in a given linked list */
    function printList(node) {
        while (node != null) {
            document.write(node.data + " ");
            node = node.next;
        }
    }
 
     
        var k = 3; // Number of linked lists
        var n = 4; // Number of elements in each list
 
        // an array of pointers storing the head nodes
        // of the linked lists
var arr = Array(k);
 
        arr[0] = new Node(1);
        arr[0].next = new Node(3);
        arr[0].next.next = new Node(5);
        arr[0].next.next.next = new Node(7);
 
        arr[1] = new Node(2);
        arr[1].next = new Node(4);
        arr[1].next.next = new Node(6);
        arr[1].next.next.next = new Node(8);
 
        arr[2] = new Node(0);
        arr[2].next = new Node(9);
        arr[2].next.next = new Node(10);
        arr[2].next.next.next = new Node(11);
 
        // Merge all lists
var head = mergeKLists(arr, k - 1);
        printList(head);
 
// This code contributed by gauravrajput1
 

输出：

0 1 2 3 4 5 6 7 8 9 10 11

复杂度分析：

假设N(n*k)是节点总数，n是每个链表的大小，k是链表的总数。

• 时间复杂度： O(N*log k) 或 O(n*k*log k)
  作为函数mergeKLists() 中的外部 while 循环运行 log k 次，并且每次它处理 n*k 个元素。
• 辅助空间： O(N) 或 O(n*k)
  因为在 SortedMerge() 中使用递归并合并大小为 N/2 的最后 2 个链表，所以将进行 N 次递归调用。

合并k个排序链表|设置 2（使用最小堆）

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