main.cpp

// Source: https://leetcode.com/problems/reverse-linked-list
// Title: Reverse Linked List
// Difficulty: Easy
// Author: Mu Yang <http://muyang.pro>

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// Given the `head` of a singly linked list, reverse the list, and return the reversed list.
//
// **Example 1:**
// https://assets.leetcode.com/uploads/2021/02/19/rev1ex1.jpg
//
// ```
// Input: head = [1,2,3,4,5]
// Output: [5,4,3,2,1]
// ```
//
// **Example 2:**
// https://assets.leetcode.com/uploads/2021/02/19/rev1ex2.jpg
//
// ```
// Input: head = [1,2]
// Output: [2,1]
// ```
//
// **Example 3:**
//
// ```
// Input: head = []
// Output: []
// ```
//
// **Constraints:**
//
// - The number of nodes in the list is the range `[0, 5000]`.
// - `-5000 <= Node.val <= 5000`
//
// **Follow up:** A linked list can be reversed either iteratively or recursively. Could you implement both?
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

using namespace std;

struct ListNode {
  int val;
  ListNode* next;
  ListNode() : val(0), next(nullptr) {}
  ListNode(int x) : val(x), next(nullptr) {}
  ListNode(int x, ListNode* next) : val(x), next(next) {}
};

class Solution {
 public:
  ListNode* reverseList(ListNode* head) {
    if (head == nullptr || head->next == nullptr) return head;

    auto prev = head, curr = prev->next;
    prev->next = nullptr;
    while (curr) {
      auto next = curr->next;
      curr->next = prev;
      prev = curr, curr = next;
    }

    return prev;
  }
};