Design a module to create a linked list form already given
linked list.New linked list must have all the even
elements of the existing list.
Answers
// 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;
}