Print Reverse a linked list using Stack

Given a linked list, print reverse of it without modifying the list.

Examples:

Input : 1 2 3 4 5 6
Output : 6 5 4 3 2 1

Input : 12 23 34 45 56 67 78
Output : 78 67 56 45 34 23 12

Given a Linked List, display the linked list in reverse without using recursion, stack or modifications to given list.



Examples:

Input : 1->2->3->4->5->NULL
Output :5->4->3->2->1->NULL

Input :10->5->15->20->24->NULL
Output :24->20->15->5->10->NULL
          

Below are different solutions that are now allowed here as we cannot use extra space and modify list.

1) Recursive solution to print reverse a linked list. Requires extra space.



2) Reverse linked list and then print. This requires modifications to original list.

3) A O(n2) solution to print reverse of linked list that first count nodes and then prints k-th node from end.

In this post, an efficient stack based solution is discussed.
1. First insert all the element in stack
2. Print stack till stack is not empty


Note: Instead of inserting data from each node into the stack, insert the node’s address onto the stack. This is because the size of the node’s data will be generally more than the size of the node’s address. Thus the stack would end up requiring more memory if it directly stored the data elements. Also, we cannot insert the node’s data onto the stack if each node contained more than one data member. Hence the simpler and efficient solution would be to simply insert the node’s address.

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// C/C++ program to print reverse of linked list
// using stack.
#include<bits/stdc++.h>
using namespace std;
   
/* Link list node */
struct Node
{
    int data;
    struct Node* next;
};
   
/* Given a reference (pointer to pointer) to the head
  of a list and an int, push a new node on the front
  of the list. */
void push(struct Node** head_ref, int new_data)
{
    struct Node* new_node =
            (struct Node*) malloc(sizeof(struct Node));
    new_node->data  = new_data;
    new_node->next = (*head_ref);
    (*head_ref)    = new_node;
}
   
/* Counts no. of nodes in linked list */
int getCount(struct Node* head)
{
    int count = 0;  // Initialize count
    struct Node* current = head;  // Initialize current
    while (current != NULL)
    {
        count++;
        current = current->next;
    }
    return count;
}
   
/* Takes head pointer of the linked list and index
    as arguments and return data at index*/
int getNth(struct Node* head, int n)
{
    struct Node* curr = head;
    for (int i=0; i<n-1 && curr != NULL; i++)
       curr = curr->next;
    return curr->data;
}
   
void printReverse(Node *head)
{
    // store Node addresses in stack
    stack<Node *> stk; 
    Node* ptr = head;
    while (ptr != NULL) 
    {
        stk.push(ptr);
        ptr = ptr->next;
    }
  
    // print data from stack
    while (!stk.empty())
    {
        cout << stk.top()->data << " ";
        stk.pop(); // pop after print
    }
    cout << "\n";
}
   
/* Drier program to test count function*/
int main()
{
    /* Start with the empty list */
    struct Node* head = NULL;
   
    /* Use push() to construct below list
     1->2->3->4->5 */
    push(&head, 5);
    push(&head, 4);
    push(&head, 3);
    push(&head, 2);
    push(&head, 1);
   
    printReverse(head);
   
    return 0;
}

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Output:

5 4 3 2 1 


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Improved By : Sayan Mahapatra