Given two linked lists, insert nodes of second list into first list at alternate positions of first list.
For example, if first list is 5->7->17->13->11 and second is 12->10->2->4->6, the first list should become 5->12->7->10->17->2->13->4->11->6 and second list should become empty. The nodes of second list should only be inserted when there are positions available. For example, if the first list is 1->2->3 and second list is 4->5->6->7->8, then first list should become 1->4->2->5->3->6 and second list to 7->8.
Use of extra space is not allowed (Not allowed to create additional nodes), i.e., insertion must be done in-place. Expected time complexity is O(n) where n is number of nodes in first list.
The idea is to run a loop while there are available positions in first loop and insert nodes of second list by changing pointers. Following are implementations of this approach.
C++
// C++ program to merge a linked list // into another at alternate positions #include <bits/stdc++.h>using namespace std;// A nexted list node class Node { public: int data; Node *next; }; /* Function to insert a node at the beginning */void push(Node ** head_ref, int new_data) { Node* new_node = new Node(); new_node->data = new_data; new_node->next = (*head_ref); (*head_ref) = new_node; } /* Utility function to print a singly linked list */void printList(Node *head) { Node *temp = head; while (temp != NULL) { cout << temp -> data << " "; temp = temp -> next; } cout << endl;} // Main function that inserts nodes of // linked list q into p at alternate positions. // Since head of first list never changes and // head of second list may change, we need single // pointer for first list and double pointer for // second list. void merge(Node *p, Node **q) { Node *p_curr = p, *q_curr = *q; Node *p_next, *q_next; // While there are available positions // in p while (p_curr != NULL && q_curr != NULL) { // Save next pointers p_next = p_curr->next; q_next = q_curr->next; // Make q_curr as next of p_curr // Change next pointer of q_curr q_curr->next = p_next; // Change next pointer of p_curr p_curr->next = q_curr; // Update current pointers for // next iteration p_curr = p_next; q_curr = q_next; } // Update head pointer of second list *q = q_curr; } // Driver code int main() { Node *p = NULL, *q = NULL; push(&p, 3); push(&p, 2); push(&p, 1); cout << "First Linked List:"; printList(p); push(&q, 8); push(&q, 7); push(&q, 6); push(&q, 5); push(&q, 4); cout << "Second Linked List:"; printList(q); merge(p, &q); cout << "Modified First Linked List:"; printList(p); cout << "Modified Second Linked List:"; printList(q); return 0; } // This code is contributed by rathbhupendra |
Output:
First Linked List: 1 2 3 Second Linked List: 4 5 6 7 8 Modified First Linked List: 1 4 2 5 3 6 Modified Second Linked List: 7 8
Time Complexity: O(min(n1, n2)), where n1 and n2 represents the length of the given two linked lists.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Please refer complete article on Merge a linked list into another linked list at alternate positions for more details!
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