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, & next 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) {}
 * };
*/
classSolution {
public:
 ListNode* reverseList(ListNode* head) {
if (head == NULL || head->next == NULL) {
return head;
 }

 ListNode* prev = NULL;
 ListNode* curr = head;
 ListNode* next = curr->next;

while (curr != NULL) {
 next = curr->next;
 curr->next = prev;
 prev = curr;
 curr = next;
 }

return prev;
 }
};