#206

Reverse Linked List

easy· Linked Listruns: 0

Given the head of a singly linked list, reverse the list and return the new head.

wpm 0acc 100%time 0:000 / 281

class Solution:
    def reverseList(self, head: ListNode | None) -> ListNode | None:
        prev = None
        curr = head
        while curr:
            next_node = curr.next
            curr.next = prev
            prev = curr
            curr = next_node
        return prev

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