We must print the k nodes of the linked list in reverse order. We have to apply an iterative approach to solve this problem.
Iteration methods usually use a loop to execute until the condition value is 1 or true.
Say, list contains nodes 29, 34, 43, 56 and 88 and the value of k is 2, the output will be the spare nodes up to k, for example 56 and 88.
Linked List: 29->34->43->56->88 Input: 2 Output: 56 88
Since we have to remove the last k elements from the list, the best way is to use a stack data structure where the elements are pushed into it, this creates the list and the starting element of the stack is the last element of the list and then they are popped off the stack until the kth time, giving us the last node of the linked list.
The code below shows the C implementation of the given algorithm.
START Step 1 -> create node variable of type structure Declare int data Declare pointer of type node using *next Step 2 -> create struct node* intoList(int data) Create newnode using malloc Set newnode->data = data newnode->next = NULL return newnode step 3 -> Declare function void rev(struct node* head,int count, int k) create struct node* temp1 = head Loop While(temp1 != NULL) count++ temp1 = temp1->next end Declare int array[count], temp2 = count,i Set temp1 = head Loop While(temp1 != NULL) Set array[--temp2] = temp1->data Set temp1 = temp1->next End Loop For i = 0 and i < k and i++ Print array[i] End Step 4 -> In Main() Create list using struct node* head = intoList(9) Set k=3 and count=0 Call rev(head,count,k) STOP
#include<stdio.h> #include<stdlib.h> // Structure of a node struct node { int data; struct node *next; }; //functon for inserting a new node struct node* intoList(int data) { struct node* newnode = (struct node*)malloc(sizeof(struct node)); newnode->data = data; newnode->next = NULL; return newnode; } // Function to reversely printing the elements of a node void rev(struct node* head,int count, int k) { struct node* temp1 = head; while(temp1 != NULL) { count++; temp1 = temp1->next; } int array[count], temp2 = count,i; temp1 = head; while(temp1 != NULL) { array[--temp2] = temp1->data; temp1 = temp1->next; } for(i = 0; i < k; i++) printf("%d ",array[i]); } int main() { printf("</p><p>reverse of a list is : "); struct node* head = intoList(9); //inserting elements into a list head->next = intoList(76); head->next->next = intoList(13); head->next->next->next = intoList(24); head->next->next->next->next = intoList(55); head->next->next->next->next->next = intoList(109); int k = 3, count = 0; rev(head, count, k); //calling function to print reversely return 0; }
If we run the above program, it will generate the following output.
reverse of a list is : 109 55 24
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