Question

Arrays, Lists, Stacks and Queues:

1) Write C++ code to reverse a singly-linked list L using only a constant amount of additional storage and no recursion. Assume the list object has a head pointer _head and consists of Node objects; a Node object has a pointer Node* _next

Answer 1

Code Screenshot:

Code Output:

Code:

#include<bits/stdc++.h>
using namespace std;
// Structure to store a linked list Node
struct Node
{
int value;
struct Node *_next;
};

void
push_front (struct Node **_head, int value)
{
struct Node *new_node = (struct Node *) malloc (sizeof (struct Node));
new_node->value = value;
new_node->_next = *_head;
*_head = new_node;
}

// reverse linked list iteratively
void reverse_list (struct Node **_head)
{
struct Node *previous = NULL;
struct Node *current = *_head;
// iterate through the list
while (current != NULL)
{
// store next of current in next
struct Node *next = current->_next;
// move the current node onto the previous
current->_next = previous;
previous = current;
// process next node
current = next;
}
// set head pointer
*_head = previous;
}

int main ()
{
int n;
cout << "Enter no of node in linked list";
cin >> n;
int i, value;
struct Node *_head = NULL;
for (i = 0; i < n; i++)
{
cin >> value;
push_front (&_head, value);
  
}
cout << "Linked list Initially:\n";
struct Node *temp = _head;
while (temp)
{
cout<< temp->value<<"->";
temp = temp->_next;
}
cout<<"Null";
reverse_list (&_head);
cout << "\nLinked list after reverse:\n";

temp = _head;
while (temp)
{
cout<< temp->value<<"->";
temp = temp->_next;
}
cout<<"Null";

return 0;
}

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