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

 

Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both?





 # 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_node = None
        current_node = head
        while current_node is not None:
            temp_node = current_node.next
            current_node.next = prev_node
            prev_node = current_node
            current_node = temp_node

        return prev_node

Random Note


when you have something circular, most of the time you can have one variableand another can be calculated using size, current value and doing %