Computer Science, asked by shobhanshubaibil, 9 months ago

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

Answered by Anonymous
2

// 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;

}

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