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Reverse Linked List

Question​

Given a singly linked list, reverse the order of the nodes in the linked list, and return the head of the reversed linked list.

Example 1
Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL

Solution​

all//Reverse Linked List.py


# Definition for singly-linked list.
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        # Initialize the prev and current node
        prev = None
        curr = head

        # Loop through the list
        while curr:
            # Store the next node
            next_node = curr.next
            # Reverse the link by pointing current node to the prev node
            curr.next = prev
            # Update the prev node and current node
            prev = curr
            curr = next_node
        # Return the prev node which is the head of the reversed list
        return prev