Given the head of a singly linked list, reverse the list in place and return the new head. Each node contains an integer value and a next pointer. If the list is empty, return None.
Constraints
0 <= n <= 5000, where n is the number of nodes
-5000 <= Node.val <= 5000
The list is singly linked
The solution should run in linear time
Function Signature
defreverse_linked_list(head):
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