Given a Singly Linked List, write a function to delete a given node. Your function must follow the following constraints:
- It must accept a pointer to the start node as the first parameter and node to be deleted as the second parameter i.e., a pointer to the head node is not global.
- It should not return a pointer to the head node.
- It should not accept pointer to pointer to the head node.
You may assume that the Linked List never becomes empty.
Let the function name be deleteNode(). In a straightforward implementation, the function needs to modify the head pointer when the node to be deleted is the first node. As discussed in previous post, when a function modifies the head pointer, the function must use one of the given approaches, we can’t use any of those approaches here.
Solution:
We explicitly handle the case when the node to be deleted is the first node, we copy the data of the next node to head and delete the next node. The cases when a deleted node is not the head node can be handled normally by finding the previous node and changing the next of the previous node. The following are the implementation.
C++
// C++ program to delete a given node// in linked list under given constraints#include <bits/stdc++.h>using namespace std;// Structure of a linked list // node class Node { public: int data; Node *next; }; void deleteNode(Node *head, Node *n) { // When node to be deleted is // head node if(head == n) { if(head->next == NULL) { cout << "There is only one node." << " The list can't be made empty "; return; } // Copy the data of next node // to head head->data = head->next->data; // Store address of next node n = head->next; // Remove the link of next node head->next = head->next->next; // Free memory free(n); return; } // When not first node, follow // the normal deletion process // Find the previous node Node *prev = head; while(prev->next != NULL && prev->next != n) prev = prev->next; // Check if node really exists in // Linked List if(prev->next == NULL) { cout << "Given node is not present in Linked List"; return; } // Remove node from Linked List prev->next = prev->next->next; // Free memory free(n); return; } /* Utility 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 linked list */void printList(Node *head) { while(head != NULL) { cout << head->data << " "; head = head->next; } cout << endl;} // Driver codeint main() { Node *head = NULL; /* Create following linked list 12->15->10->11->5->6->2->3 */ push(&head,3); push(&head,2); push(&head,6); push(&head,5); push(&head,11); push(&head,10); push(&head,15); push(&head,12); cout << "Given Linked List: "; printList(head); /* Let us delete the node with value 10 */ cout << "Deleting node " << head->next->next->data << " "; deleteNode(head, head->next->next); cout << "Modified Linked List: "; printList(head); // Let us delete the first node cout << "Deleting first node "; deleteNode(head, head); cout << "Modified Linked List: "; printList(head); return 0; } // This code is contributed by rathbhupendra |
Output:
Given Linked List: 12 15 10 11 5 6 2 3 Deleting node 10: Modified Linked List: 12 15 11 5 6 2 3 Deleting first node Modified Linked List: 15 11 5 6 2 3
Time Complexity: O(n), where n represents the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Please refer complete article on Delete a given node in Linked List under given constraints for more details!
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