The correct, optimal and working solution for programming question reverse-linked-list on leetcode

/* * Author: Arpit Bhayani * https://arpitbhayani.me */ #include <cmath> #include <cstdio> #include <cstdlib> #include <climits> #include <deque> #include <iostream> #include <list> #include <limits> #include <map> #include <queue> #include <set> #include <stack> #include <vector> #define ll long long #define MIN(a, b) a < b ? a : b #define MAX(a, b) a > b ? a : b using namespace std; int readline(char *str) { int i = 0; char ch; while((ch = getchar()) != '\n') { str[i++] = ch; } str[i] = '\0'; return i; } struct node { int val; struct node * next; }; struct node * new_node(int d) { struct node * t = (struct node *) calloc(1, sizeof(struct node)); t->val = d; return t; } void printList(struct node * head) { struct node * p = head; while(p) { printf("%d ", p->val); p = p->next; } printf("\n"); } struct node * reverse_list(struct node * head) { struct node * p = head; struct node * q = NULL; struct node * r = NULL; while(p) { r = p->next; p->next = q; q = p; p = r; } return q; } int main(int argc, char *argv[]) { struct node * head = new_node(2); head->next = new_node(1); head->next->next = new_node(2); head->next->next->next = new_node(4); head->next->next->next->next = new_node(5); head->next->next->next->next->next = new_node(6); printList(head); head = reverse_list(head); printList(head); return 0; }