Constraints
- The number of nodes in the list is the range [0, 5000].
- -5000 <= Node.val <= 5000
Idea #1 (Time: N^2, Memory: N)
- until End of List 1.1. pop 1.2. appendleft
Idea #2 (Time: N^2, Memory: N)
- until End of List 1.1. popleft 1.2. append
Test Cases
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: []
Code
class ListNode: def __init__(self, val=0, next=None): self.val = val self.next = next class LinkedList: def __init__(self, head=None): self.head = head def reverse(self): if self.head == None or self.head.next == None: return prev = None ahead = self.head.next while ahead: self.head.next = prev prev = self.head self.head = ahead ahead = ahead.next self.head.next = prev class Solution: def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]: linkedlist = LinkedList(head) linkedlist.reverse() return linkedlist.head 
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