Delete last occurrence of an item from linked list
Last Updated : 20 Sep, 2024
Given a singly linked list and a key, the task is to delete the last occurrence of that key in the linked list.
Examples:
Input: head: 1 -> 2 -> 3 ->1 -> 4 -> NULL, key = 1
Output: 1 -> 2 -> 3 -> 4 -> NULL
Input: head: 1 -> 2 -> 3 -> 4 -> 5 -> NULL , key = 3
Output: 1 -> 2 -> 4 -> 5 -> NULL
Approach:
The idea is to traverse the linked list from beginning to end. While traversing, keep track of last occurrence key node and previous node of that key. After traversing the complete list, delete the last occurrence of that key.
Follow the steps below to solve the problem:
- Initialize Pointer curr points to head, last, lastPrev and prev to NULL.
- Traverse the List until curr is not NULL:
- If curr->data == key, update lastPrev to the prev and last to curr.
- Move prev pointer to curr and curr to the next node.
- Delete Last Occurrence (if key was found then last is not null):
- If lastPrev is not null, adjust lastPrev->next = last->next to skip last.
- If last is the head, update the head to last->next.
C++ // C++ program to delete last occurrence // of key in singly linked list #include <iostream> using namespace std; class Node { public: int data; Node* next; Node(int x) { data = x; next = nullptr; } }; // Function to delete the last occurrence // of a key in the linked list Node* deleteLastOccurrence(Node* head, int key) { Node *last = nullptr, *lastPrev = nullptr; Node *curr = head, *prev = nullptr; // Traverse the list to find the last // occurrence of the key while (curr != nullptr) { if (curr->data == key) { lastPrev = prev; last = curr; } prev = curr; curr = curr->next; } // If the key was found if (last != nullptr) { // If last occurrence is not the head if (lastPrev != nullptr) { lastPrev->next = last->next; } else { // If last occurrence is the head, // move head to next node head = head->next; } delete last; } return head; } void print(Node* curr) { while (curr != nullptr) { cout << curr->data << " "; curr = curr->next; } cout << endl; } int main() { // Create a hard-coded linked list: // 1 -> 2 -> 2 -> 4 -> 2 Node* head = new Node(1); head->next = new Node(2); head->next->next = new Node(2); head->next->next->next = new Node(4); head->next->next->next->next = new Node(2); int key = 2; head = deleteLastOccurrence(head, key); print(head); return 0; } // C program to delete last occurrence // of key in singly linked list #include <stdio.h> #include <stdlib.h> struct Node { int data; struct Node* next; }; // Function to delete the last occurrence // of a key in the linked list struct Node* deleteLastOccurrence(struct Node* head, int key) { struct Node *last = NULL, *lastPrev = NULL; struct Node *curr = head, *prev = NULL; // Traverse the list to find the last // occurrence of the key while (curr != NULL) { if (curr->data == key) { lastPrev = prev; last = curr; } prev = curr; curr = curr->next; } // If the key was found if (last != NULL) { // If last occurrence is not the head if (lastPrev != NULL) { lastPrev->next = last->next; } else { // If last occurrence is the head, // move head to next node head = head->next; } free(last); } return head; } void print(struct Node* curr) { while (curr != NULL) { printf("%d ", curr->data); curr = curr->next; } printf("\n"); } struct Node* createNode(int new_data) { struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); newNode->data = new_data; newNode->next = NULL; return newNode; } int main() { // Create a hard-coded linked list: // 1 -> 2 -> 2 -> 4 -> 2 struct Node* head = createNode(1); head->next = createNode(2); head->next->next = createNode(2); head->next->next->next = createNode(4); head->next->next->next->next = createNode(2); int key = 2; head = deleteLastOccurrence(head, key); print(head); return 0; } // Java program to delete last occurrence // of key in singly linked list class Node { int data; Node next; Node(int x) { data = x; next = null; } } class GfG { // Function to delete the last occurrence // of a key in the linked list static Node deleteLastOccurrence(Node head, int key) { Node last = null, lastPrev = null; Node curr = head, prev = null; // Traverse the list to find the last // occurrence of the key while (curr != null) { if (curr.data == key) { lastPrev = prev; last = curr; } prev = curr; curr = curr.next; } // If the key was found if (last != null) { // If last occurrence is not the head if (lastPrev != null) { lastPrev.next = last.next; } else { // If last occurrence is the head, // move head to next node head = head.next; } } return head; } static void print(Node curr) { while (curr != null) { System.out.print(curr.data + " "); curr = curr.next; } System.out.println(); } public static void main(String[] args) { // Create a hard-coded linked list: // 1 -> 2 -> 2 -> 4 -> 2 Node head = new Node(1); head.next = new Node(2); head.next.next = new Node(2); head.next.next.next = new Node(4); head.next.next.next.next = new Node(2); int key = 2; head = deleteLastOccurrence(head, key); print(head); } } # Python program to delete last occurrence # of key in singly linked list class Node: def __init__(self, x): self.data = x self.next = None # Function to delete the last occurrence # of a key in the linked list def deleteLastOccurrence(head, key): last = None lastPrev = None curr = head prev = None # Traverse the list to find the last # occurrence of the key while curr is not None: if curr.data == key: lastPrev = prev last = curr prev = curr curr = curr.next # If the key was found if last is not None: # If last occurrence is not the head if lastPrev is not None: lastPrev.next = last.next else: # If last occurrence is the head, # move head to next node head = head.next return head def printList(curr): while curr is not None: print(curr.data, end=" ") curr = curr.next print() if __name__ == "__main__": # Create a hard-coded linked list: # 1 -> 2 -> 2 -> 4 -> 2 head = Node(1) head.next = Node(2) head.next.next = Node(2) head.next.next.next = Node(4) head.next.next.next.next = Node(2) key = 2 head = deleteLastOccurrence(head, key) printList(head)
// C# program to delete last occurrence // of key in singly linked list class Node { public int data; public Node next; public Node(int new_data) { data = new_data; next = null; } } class GfG { // Function to delete the last occurrence // of a key in the linked list static Node deleteLastOccurrence(Node head, int key) { Node last = null, lastPrev = null; Node curr = head, prev = null; // Traverse the list to find the last // occurrence of the key while (curr != null) { if (curr.data == key) { lastPrev = prev; last = curr; } prev = curr; curr = curr.next; } // If the key was found if (last != null) { // If last occurrence is not the head if (lastPrev != null) { lastPrev.next = last.next; } else { // If last occurrence is the head, // move head to next node head = head.next; } } return head; } static void print(Node curr) { while (curr != null) { System.Console.Write(curr.data + " "); curr = curr.next; } System.Console.WriteLine(); } static void Main(string[] args) { // Create a hard-coded linked list: // 1 -> 2 -> 2 -> 4 -> 2 Node head = new Node(1); head.next = new Node(2); head.next.next = new Node(2); head.next.next.next = new Node(4); head.next.next.next.next = new Node(2); int key = 2; head = deleteLastOccurrence(head, key); print(head); } } // Javascript program to delete last occurrence // of key in singly linked list class Node { constructor(new_data) { this.data = new_data; this.next = null; } } // Function to delete the last occurrence // of a key in the linked list function deleteLastOccurrence(head, key) { let last = null, lastPrev = null; let curr = head, prev = null; // Traverse the list to find the last // occurrence of the key while (curr !== null) { if (curr.data === key) { lastPrev = prev; last = curr; } prev = curr; curr = curr.next; } // If the key was found if (last !== null) { // If last occurrence is not the head if (lastPrev !== null) { lastPrev.next = last.next; } else { // If last occurrence is the head, // move head to next node head = head.next; } } return head; } function printList(curr) { while (curr !== null) { console.log(curr.data + " "); curr = curr.next; } } // Create a hard-coded linked list: // 1 -> 2 -> 2 -> 4 -> 2 let head = new Node(1); head.next = new Node(2); head.next.next = new Node(2); head.next.next.next = new Node(4); head.next.next.next.next = new Node(2); let key = 2; head = deleteLastOccurrence(head, key); printList(head); Time Complexity: O(n), Traversing over the linked list of size n.
Auxiliary Space: O(1)