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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
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