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 from 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.
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!
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!