C++ Program To Merge K Sorted Linked Lists - Set 1 - GeeksforGeeks

03 Jan, 2022
Given K sorted linked lists of size N each, merge them and print the sorted output.
Examples: 
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.
Method 1 (Simple):
Approach: A Simple Solution is to initialize the result as the first list. Now traverse all lists starting from the second list. Insert every node of the currently traversed list into result in a sorted way.  
C++
// C++ program to merge k sorted// linked lists of size n each#include <bits/stdc++.h>using namespace std;  // A Linked List nodestruct 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 outputNode* 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 codeint 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;}
Output:
0 1 2 3 4 5 6 7 8 9 10 11
Complexity Analysis: 
Time complexity: O(nk2)
Auxiliary Space: O(1). As no extra space is required.
Method 2: Min Heap. A Better solution is to use Min Heap-based solution which is discussed here for arrays. The time complexity of this solution would be O(nk Log k)Method 3: Divide and Conquer. In this post, Divide and Conquer approach is discussed. This approach doesn’t require extra space for heap and works in O(nk Log k)It is known that merging of two linked lists can be done in O(n) time and O(n) space. 
The idea is to pair up K lists and merge each pair in linear time using O(n) space.After the first cycle, K/2 lists are left each of size 2*N. After the second cycle, K/4 lists are left each of size 4*N and so on.Repeat the procedure until we have only one list left.
The idea is to pair up K lists and merge each pair in linear time using O(n) space.
After the first cycle, K/2 lists are left each of size 2*N. After the second cycle, K/4 lists are left each of size 4*N and so on.
Repeat the procedure until we have only one list left.
Below is the implementation of the above idea. 
C++
// C++ program to merge k sorted// linked lists of size n each#include <bits/stdc++.h>using namespace std;  // A Linked List nodestruct 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 outputNode* 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 codeint 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;}
Output:
0 1 2 3 4 5 6 7 8 9 10 11
Complexity Analysis: 
Assuming N(n*k) is the total number of nodes, n is the size of each linked list, and k is the total number of linked lists.
Time Complexity: O(N*log k) or O(n*k*log k)As outer while loop in function mergeKLists() runs log k times and every time it processes n*k elements.
Auxiliary Space: O(N) or O(n*k)Because recursion is used in SortedMerge() and to merge the final 2 linked lists of size N/2, N recursive calls will be made.
Please refer complete article on Merge K sorted linked lists | Set 1 for more details!
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