1. Problem
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:
- The number of nodes in the list is the range [0, 5000].
- -5000 <= Node.val <= 5000
2. Solution
I solved this problem like this. Push all node in stack and make new linked list.
# 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 head
stack = []
node = head
while node:
stack.append(node)
node = node.next
stack[-1].next = None
new_head = stack[-1]
stack.pop()
node = new_head
while stack:
node.next = stack[-1]
node = node.next
stack.pop()
return new_head
Other solution is like this. This solution is simple and easy to understand.
def reverseList(self, head):
new_list = None
current = head
while current:
next_node = current.next
current.next = new_list
new_list = current
current = next_node
return new_list