head of a singly linked list, reverse the list, and return the reversed list.
Input: head = [1,2,3,4,5] Output: [5,4,3,2,1]
Input: head = [1,2] Output: [2,1]
Input: head =  Output: 
- The number of nodes in the list is the range
-5000 <= Node.val <= 5000
Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both?
Keep track of the previous nodes. Attach each node’s next to the previous node.
# Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]: prev = None while head != None: temp = head.next head.next = prev prev = head head = temp return prev
- Time complexity: O(N).
- Space complexity: O(1).