Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Print reverse of a Linked List without extra space and modification in C Program.
The task is to print the nodes starting from the end of the linked list without using extra space which means there shouldn’t be any extra variable instead the head pointer pointing the first node will be moved.
Example
Input: 10 21 33 42 89 Output: 89 42 33 21 10
There can be many solutions to print the linked list in reverse order, like recursive approach (use extra space), reverse the linked list(requires modification in the given Linked List), pushing the elements on a stack and then pop and display the elements one by one(requires space O(n)), but these solutions seem to be using more space than the O(1).
To achieve the result without using more than O(1) we can −
- Count the number of nodes in a linked list
- Loop from i = n to 1 and print the i-th location’s node.
Algorithm
START Step 1 -> create node variable of type structure Declare int data Declare pointer of type node using *next Step 2 ->Declare function int get(struct node* head) Declare variable as int count=0 Declare struct node *newme=head Loop While newme!=NULL Increment count by 1 Set newme = newme->next End Return count Step 3 -> Declare Function void push(node** headref, char newdata) Allocate memory using malloc Set newnode->data = newdata Set newnode->next = (*headref) Set (*headref) = newnode Step 4 -> Declare function int getN(struct node* head, int n) Declare struct node* cur = head Loop for int i=0 and i<n-1 && cur != NULL and i++ Set cur=cur->next End Return cur->dataStep 5 -> Declare function void reverse(node *head) Declare int n = get(head) Loop For int i=n and i>=1 and i— Print getN(head,i) End Step 6 ->In Main() Create list using node* head = NULL Insert elements through push(&head, 89) Call reverse(head) STOP
Example
#include<stdio.h>
#include<stdlib.h>
//node structure
struct node {
int data;
struct node* next;
};
void push(struct node** headref, int newdata) {
struct node* newnode = (struct node*) malloc(sizeof(struct node));
newnode->data = newdata;
newnode->next = (*headref);
(*headref) = newnode;
}
int get(struct node* head) {
int count = 0;
struct node* newme = head;
while (newme != NULL){
count++;
newme = newme->next;
}
return count;
}
int getN(struct node* head, int n) {
struct node* cur = head;
for (int i=0; i<n-1 && cur != NULL; i++)
cur = cur->next;
return cur->data;
}
void reverse(node *head) {
int n = get(head);
for (int i=n; i>=1; i--)
printf("%d ", getN(head, i));
}
int main() {
struct node* head = NULL; //create a first node
push(&head, 89); //pushing element in the list
push(&head, 42);
push(&head, 33);
push(&head, 21);
push(&head, 10);
reverse(head); //calling reverse function
return 0;
}Output
if we run above program then it will generate following output
89 42 33 21 10
Updated on: 22-Aug-2019
313 Views
Advertisements