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Create a linked list from two linked lists by choosing max element at each position in C++ Program
In this tutorial, we are going to write a program that creates a new linked list from the given linked lists.
We have given two linked lists of the same size and we have to create a new linked list from the two linked lists with the max numbers from the two linked lists.
Let's see the steps to solve the problem.
Write a struct node.
Create two linked lists of the same size.
Iterate over the linked list.
Find the max number from the two linked lists nodes.
Create a new node with the max number.
Add the new node to the new linked list.
Print the new linked list.
Example
Let's see the code.
#include <bits/stdc++.h>
using namespace std;
struct Node {
int data;
Node* next;
};
void insertNewNode(Node** root, int item) {
Node *ptr, *temp;
temp = new Node;
temp->data = item;
temp->next = NULL;
if (*root == NULL) {
*root = temp;
}
else {
ptr = *root;
while (ptr->next != NULL) {
ptr = ptr->next;
}
ptr->next = temp;
}
}
void printLinkedList(Node* root) {
while (root != NULL) {
cout << root->data << " -> ";
root = root->next;
}
cout << "NULL" << endl;
}
Node* generateNewLinkedList(Node* root1, Node* root2) {
Node *ptr1 = root1, *ptr2 = root2;
Node* root = NULL;
while (ptr1 != NULL) {
int currentMax = ((ptr1->data < ptr2->data) ? ptr2->data : ptr1->data);
if (root == NULL) {
Node* temp = new Node;
temp->data = currentMax;
temp->next = NULL;
root = temp;
}
else {
insertNewNode(&root, currentMax);
}
ptr1 = ptr1->next;
ptr2 = ptr2->next;
}
return root;
}
int main() {
Node *root1 = NULL, *root2 = NULL, *root = NULL;
insertNewNode(&root1, 1);
insertNewNode(&root1, 2);
insertNewNode(&root1, 3);
insertNewNode(&root1, 4);
insertNewNode(&root2, 3);
insertNewNode(&root2, 1);
insertNewNode(&root2, 2);
insertNewNode(&root2, 4);
root = generateNewLinkedList(root1, root2);
printLinkedList(root);
return 0;
}Output
If you run the above code, then you will get the following result.
3 -> 2 -> 3 -> 4 -> NULL
Conclusion
If you have any queries in the tutorial, mention them in the comment section.
Updated on: 28-Jan-2021
2K+ Views
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