// A simple CPP program to introduce
// a linked list
#include <bits/stdc++.h>
using namespace std;
class Node {
public:
int data;
Node* next;
};
// Program to create a simple linked
// list with 3 nodes
int main()
{
Node* head = NULL;
Node* second = NULL;
Node* third = NULL;
// allocate 3 nodes in the heap
head = new Node();
second = new Node();
third = new Node();
/* Three blocks have been allocated dynamically.
We have pointers to these three blocks as head,
second and third
head second third
| | |
| | |
+---+-----+ +----+----+ +----+----+
| # | # | | # | # | | # | # |
+---+-----+ +----+----+ +----+----+
# represents any random value.
Data is random because we haven’t assigned
anything yet */
head->data = 1; // assign data in first node
head->next = second; // Link first node with
// the second node
/* data has been assigned to the data part of first
block (block pointed by the head). And next
pointer of the first block points to second.
So they both are linked.
head second third
| | |
| | |
+---+---+ +----+----+ +-----+----+
| 1 | o----->| # | # | | # | # |
+---+---+ +----+----+ +-----+----+
*/
// assign data to second node
second->data = 2;
// Link second node with the third node
second->next = third;
/* data has been assigned to the data part of the second
block (block pointed by second). And next
pointer of the second block points to the third
block. So all three blocks are linked.
head second third
| | |
| | |
+---+---+ +---+---+ +----+----+
| 1 | o----->| 2 | o-----> | # | # |
+---+---+ +---+---+ +----+----+ */
third->data = 3; // assign data to third node
third->next = NULL;
/* data has been assigned to the data part of the third
block (block pointed by third). And next pointer
of the third block is made NULL to indicate
that the linked list is terminated here.
We have the linked list ready.
head
|
|
+---+---+ +---+---+ +----+------+
| 1 | o----->| 2 | o-----> | 3 | NULL |
+---+---+ +---+---+ +----+------+
Note that only the head is sufficient to represent
the whole list. We can traverse the complete
list by following the next pointers. */
return 0;
}
// This code is contributed by rathbhupendra// A simple C program to introduce
// a linked list
#include <stdio.h>
#include <stdlib.h>
struct Node {
int data;
struct Node* next;
};
// Program to create a simple linked
// list with 3 nodes
int main()
{
struct Node* head = NULL;
struct Node* second = NULL;
struct Node* third = NULL;
// allocate 3 nodes in the heap
head = (struct Node*)malloc(sizeof(struct Node));
second = (struct Node*)malloc(sizeof(struct Node));
third = (struct Node*)malloc(sizeof(struct Node));
/* Three blocks have been allocated dynamically.
We have pointers to these three blocks as head,
second and third
head second third
| | |
| | |
+---+-----+ +----+----+ +----+----+
| # | # | | # | # | | # | # |
+---+-----+ +----+----+ +----+----+
# represents any random value.
Data is random because we haven’t assigned
anything yet */
head->data = 1; // assign data in first node
head->next = second; // Link first node with
// the second node
/* data has been assigned to the data part of the first
block (block pointed by the head). And next
pointer of first block points to second.
So they both are linked.
head second third
| | |
| | |
+---+---+ +----+----+ +-----+----+
| 1 | o----->| # | # | | # | # |
+---+---+ +----+----+ +-----+----+
*/
// assign data to second node
second->data = 2;
// Link second node with the third node
second->next = third;
/* data has been assigned to the data part of the second
block (block pointed by second). And next
pointer of the second block points to the third
block. So all three blocks are linked.
head second third
| | |
| | |
+---+---+ +---+---+ +----+----+
| 1 | o----->| 2 | o-----> | # | # |
+---+---+ +---+---+ +----+----+ */
third->data = 3; // assign data to third node
third->next = NULL;
/* data has been assigned to data part of third
block (block pointed by third). And next pointer
of the third block is made NULL to indicate
that the linked list is terminated here.
We have the linked list ready.
head
|
|
+---+---+ +---+---+ +----+------+
| 1 | o----->| 2 | o-----> | 3 | NULL |
+---+---+ +---+---+ +----+------+
Note that only head is sufficient to represent
the whole list. We can traverse the complete
list by following next pointers. */
return 0;
}