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:
[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]: curr = head pre = None while curr: nxt = curr.next curr.next = pre pre = curr curr = nxt return pre
# 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]: if not head: return None newHead = head if head.next: newHead = self.reverseList(head.next) head.next.next = head head.next = None return newHead
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