Print nodes of linked list at given indexes in C language

We have to print the data of nodes of the linked list at the given index. Unlike array linked list generally don’t have index so we have to traverse the whole linked list and print the data when we reached a particular.

Let’s say, list contains the nodes 29, 34, 43, 56 and 88 and the value of indexes are 1, 2 and 4 than the output will be the nodes at these indexes that are 34, 43 and 88.

Example

Linked list: 29->34->43->56->88 Input: 1 2 4 Output: 34 43 88

In above representation of Linked List the yellow highlighted nodes are the nodes to be printed or the nodes which are on a particular index.

The approach used here involves taking of one pointer and one counter variable initialised to 1 that will incremented whenever the node is traversed. The counter is matched with the key value. When the key matches with the counter value the pointer pointing to the node structure will print the node’s data and incremented to next node and so on giving us the nodes at particular key.

The below code shows the c implementation of the algorithm 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 displayList(struct node *catchead)       create struct node *temp       IF catchead = NULL          Print list is empty          return       End       Set temp = catchead       Loop While (temp != NULL)          print temp->data          set temp = temp->next       End    Step 4 -> Declare Function int search(int key,struct node *head)       Set int index       Create struct node *newnode       Set index = 0 and newnode = head       Loop While (newnode != NULL & newnode->data != key)          Set index++          Set newnode = newnode->next       End       return (newnode != NULL) ? index : -1    step 5 -> In Main()       create node using struct node* head = intoList(9)       call displayList(head)       set index = search(24,head)       IF (index >= 0)          Print index       Else          Print not found in the list       EndIF STOP

Example

#include <stdio.h> #include <stdlib.h> //structure of a node struct node {    int data;    struct node *next; }; struct node* intoList(int data) {    struct node* newnode = (struct node*)malloc(sizeof(struct node));    newnode->data = data;    newnode->next = NULL;    return newnode; } //funtion to display list void displayList(struct node *catchead) {    struct node *temp;    if (catchead == NULL) {       printf("List is empty.
");       return;    }    printf("elements of list are : ");    temp = catchead;    while (temp != NULL) {       printf("%d ", temp->data);       temp = temp->next;    }    printf("
"); } //function to search element int search(int key,struct node *head) {    int index;    struct node *newnode;    index = 0;    newnode = head;    while (newnode != NULL && newnode->data != key) {       index++;       newnode = newnode->next;    }    return (newnode != NULL) ? index : -1; } int main() {    int index;    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);    displayList(head);    index = search(24,head);    if (index >= 0)       printf("%d found at position %d
", 24, index);    else       printf("%d not found in the list.
", 24);    index=search(55,head);    if (index >= 0)       printf("%d found at position %d
", 55, index);    else    printf("%d not found in the list.
", 55); }

Output

If we run above program then it will generate following output.

elements of list are : 9 76 13 24 55 109 24 found at position 3 55 found at position 4