Easy
Given the head
of a singly linked list, reverse the list, and return the reversed list.
Example 1:
Input: head = [1,2,3,4,5]
Output: [5,4,3,2,1]
Example 2:
Input: head = [1,2]
Output: [2,1]
Example 3:
Input: head = []
Output: []
Constraints:
[0, 5000]
.-5000 <= Node.val <= 5000
Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both?
/**
* 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) {
ListNode* prev = nullptr;
ListNode* curr = head;
while (curr != nullptr) {
ListNode* next = curr->next;
curr->next = prev;
prev = curr;
curr = next;
}
return prev;
}
};