Question

When you are asked to develop an algorithm, you can provide either pseudocode or actual C++ code. D 1. Assume you have a singly-linked list named L. The list header contains only one pointer, to the front of the list. Write an algorithm that displays each node starting with the first. Enter answer... 2. In words, describe an approach to displaying the list L in reverse order. Remember that L's header contains only a pointer to the first node. Enter answer...

Answer 1

#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. */
  
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.
  
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;
}