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->88Input: 1 2 4Output: 34 43 88
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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.
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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 EndIFSTOP
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Example
#include #include //structure of a nodestruct 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 listvoid 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 elementint 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);}
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输出
如果我们运行上面的程序，它将生成以下输出。
elements of list are : 9 76 13 24 55 10924 found at position 355 found at position 4
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