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


From python 3.7 dict guarantees that order will be kept as they inserted, and popitem will use LIFO order but we need FIFO type system. so we need OrderedDict which have popIten(last = T/F) for this req. One thing, next(iter(dict)) will return the first key of the dict