Se-ok JeonConstraints
- 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
