Given a liked list and a key to be deleted. Delete last occurrence of key from linked. The list may have duplicates.

Examples:

Input: 1->2->3->5->2->10, key = 2 Output: 1->2->3->5->10

The idea is to traverse the linked list from beginning to end. While traversing, keep track of last occurrence key. After traversing the complete list, delete last occurrence by copying data of next node and deleting the next node.

// A C++ program to demonstrate deletion of last // Node in singly linked list #include <bits/stdc++.h> // A linked list Node struct Node { int key; struct Node* next; }; void deleteLast(Node* head, int key) { // Initialize previous of Node to be deleted Node* x = NULL; // Start from head and find the Node to be // deleted Node* temp = head; while (temp) { // If we found the key, update xv if (temp->key == key) x = temp; temp = temp->next; } // key occurs at-least once if (x != NULL) { // Copy key of next Node to x x->key = x->next->key; // Store and unlink next temp = x->next; x->next = x->next->next; // Free memory for next delete temp; } } /* Utility function to create a new node with given key */ Node* newNode(int key) { Node* temp = new Node; temp->key = key; temp->next = NULL; return temp; } // This function prints contents of linked list // starting from the given Node void printList(struct Node* node) { while (node != NULL) { printf(" %d ", node->key); node = node->next; } } /* Drier program to test above functions*/ int main() { /* Start with the empty list */ struct Node* head = newNode(1); head->next = newNode(2); head->next->next = newNode(3); head->next->next->next = newNode(5); head->next->next->next->next = newNode(2); head->next->next->next->next->next = newNode(10); puts("Created Linked List: "); printList(head); deleteLast(head, 2); puts("\nLinked List after Deletion of 1: "); printList(head); return 0; }

Output:

Created Linked List: 1 2 3 5 2 10 Linked List after Deletion of 1: 1 2 3 5 10

**The above solution doesn’t work when the node to be deleted is the last node.**

Following solution handles all cases.

// A C program to demonstrate deletion of last // Node in singly linked list #include <stdio.h> #include <stdlib.h> // A linked list Node struct Node { int data; struct Node* next; }; // Function to delete the last occurrence void deleteLast(struct Node* head, int x) { struct Node *temp = head, *ptr = NULL; while (temp) { // If found key, update if (temp->data == x) ptr = temp; temp = temp->next; } // If the last occurrence is the last node if (ptr != NULL && ptr->next == NULL) { temp = head; while (temp->next != ptr) temp = temp->next; temp->next = NULL; } // If it is not the last node if (ptr != NULL && ptr->next != NULL) { ptr->data = ptr->next->data; temp = ptr->next; ptr->next = ptr->next->next; free(temp); } } /* Utility function to create a new node with given key */ struct Node* newNode(int x) { struct Node* node = malloc(sizeof(struct Node*)); node->data = x; node->next = NULL; return node; } // This function prints contents of linked list // starting from the given Node void display(struct Node* head) { struct Node* temp = head; if (head == NULL) { printf("NULL\n"); return; } while (temp != NULL) { printf("%d --> ", temp->data); temp = temp->next; } printf("NULL\n"); } /* Drier program to test above functions*/ int main() { struct Node* head = newNode(1); head->next = newNode(2); head->next->next = newNode(3); head->next->next->next = newNode(4); head->next->next->next->next = newNode(5); head->next->next->next->next->next = newNode(4); head->next->next->next->next->next->next = newNode(4); printf("Created Linked list: "); display(head); deleteLast(head, 4); printf("List after deletion of 4: "); display(head); return 0; }

Output:

Created Linked List: 1 2 3 4 5 4 4 Linked List after Deletion of 1: 1 2 3 4 5 4

Source:

http://qa.geeksforgeeks.org/6583/write-function-deletes-occurrence-specific-value-linked-list

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