0206-reverse-linked-list.cpp

/*
    Given the head of a singly linked list, reverse list & return
    Ex. head = [1,2,3,4,5] -> [5,4,3,2,1], head = [1,2] -> [2,1]

    Maintain prev, curr pointers, iterate thru & reverse

    Time: O(n)
    Space: O(1)
*/

/**
 * Definition for singly-linked list.
 * 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 == NULL || head->next == NULL)
            return head;

        ListNode *prev = NULL;
        ListNode *curr = head;

        while (curr != NULL)
        {
            ListNode *temp = curr->next;
            curr->next = prev;
            prev = curr;
            curr = temp;
        }
        return prev;
    }
};