Reverse Linked List 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:
- The number of nodes in the list is the range
[0, 5000]
. -5000 <= Node.val <= 5000
Reverse Linked List Solutions
✅Time: O(n)
✅Space: O(n)
C++
class Solution {
public:
ListNode* reverseList(ListNode* head) {
if (!head || !head->next)
return head;
ListNode* newHead = reverseList(head->next);
head->next->next = head;
head->next = nullptr;
return newHead;
}
};
Java
class Solution {
public ListNode reverseList(ListNode head) {
if (head == null || head.next == null)
return head;
ListNode newHead = reverseList(head.next);
head.next.next = head;
head.next = null;
return newHead;
}
}
Python
class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
if not head or not head.next:
return head
newHead = self.reverseList(head.next)
head.next.next = head
head.next = None
return newHead
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