HackerRank•Jun 12, 2025

Reverse a linked list

Hazrat Ali

HackerRank

Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty.

Example
 references the list 

Manipulate the  pointers of each node in place and return , now referencing the head of the list .

Function Description

Complete the reverse function in the editor below.

reverse has the following parameter:

  • SinglyLinkedListNode pointer head: a reference to the head of a list

Returns

  • SinglyLinkedListNode pointer: a reference to the head of the reversed list

Input Format

The first line contains an integer , the number of test cases.

Each test case has the following format:

The first line contains an integer , the number of elements in the linked list.
Each of the next  lines contains an integer, the  values of the elements in the linked list.

 

Sample Input

1
5
1
2
3
4
5

Sample Output

5 4 3 2 1 



Solution
#include <bits/stdc++.h>

using namespace std;

class SinglyLinkedListNode {
    public:
        int data;
        SinglyLinkedListNode *next;

        SinglyLinkedListNode(int node_data) {
            this->data = node_data;
            this->next = nullptr;
        }
};

class SinglyLinkedList {
    public:
        SinglyLinkedListNode *head;
        SinglyLinkedListNode *tail;

        SinglyLinkedList() {
            this->head = nullptr;
            this->tail = nullptr;
        }

        void insert_node(int node_data) {
            SinglyLinkedListNode* node = new SinglyLinkedListNode(node_data);

            if (!this->head) {
                this->head = node;
            } else {
                this->tail->next = node;
            }

            this->tail = node;
        }
};

void print_singly_linked_list(SinglyLinkedListNode* node, string sep, ofstream& fout) {
    while (node) {
        fout << node->data;

        node = node->next;

        if (node) {
            fout << sep;
        }
    }
}

void free_singly_linked_list(SinglyLinkedListNode* node) {
    while (node) {
        SinglyLinkedListNode* temp = node;
        node = node->next;

        free(temp);
    }
}

/*
 * Complete the 'reverse' function below.
 *
 * The function is expected to return an INTEGER_SINGLY_LINKED_LIST.
 * The function accepts INTEGER_SINGLY_LINKED_LIST llist as parameter.
 */

/*
 * For your reference:
 *
 * SinglyLinkedListNode {
 *     int data;
 *     SinglyLinkedListNode* next;
 * };
 *
 */

SinglyLinkedListNode* reverse(SinglyLinkedListNode* head) {
    if (!head) return head;
    SinglyLinkedListNode* prev = head;
    SinglyLinkedListNode* current = head->next;
    head->next = nullptr;

    while (current) {
        SinglyLinkedListNode* next = current->next;
        current->next = prev;
        prev = current;
        current = next;
    }

    return prev;
}

int main()
{
    ofstream fout(getenv("OUTPUT_PATH"));

    int tests;
    cin >> tests;
    cin.ignore(numeric_limits<streamsize>::max(), '\n');

    for (int tests_itr = 0; tests_itr < tests; tests_itr++) {
        SinglyLinkedList* llist = new SinglyLinkedList();

        int llist_count;
        cin >> llist_count;
        cin.ignore(numeric_limits<streamsize>::max(), '\n');

        for (int i = 0; i < llist_count; i++) {
            int llist_item;
            cin >> llist_item;
            cin.ignore(numeric_limits<streamsize>::max(), '\n');

            llist->insert_node(llist_item);
        }

        SinglyLinkedListNode* llist1 = reverse(llist->head);

        print_singly_linked_list(llist1, " ", fout);
        fout << "\n";

        free_singly_linked_list(llist1);
    }

    fout.close();

    return 0;
}






Comments