Problem
Given the head of a singly linked list, reverse the list, and return the reversed list.
https://leetcode.com/problems/reverse-linked-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
Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both?
Test Cases
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class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
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solution_test.py1
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| import pytest
from solution import Solution
from solution_recursive import Solution as SolutionRecursive
from utils import build_linked_list, format_linked_list
@pytest.mark.parametrize('head, expected', [
([1,2,3,4,5], [5,4,3,2,1]),
([1,2], [2,1]),
([], []),
])
class Test:
def test_solution(self, head, expected):
sol = Solution()
result = sol.reverseList(build_linked_list(head))
assert format_linked_list(result) == expected
def test_solution_recursive(self, head, expected):
sol = SolutionRecursive()
result = sol.reverseList(build_linked_list(head))
assert format_linked_list(result) == expected
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Thoughts
非递归版比递归版更简洁。
时间复杂度 O(n)。空间复杂度，非递归版是 O(1)，递归版是 O(n)。
Code
Iteratively
solution.py1
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| from typing import Optional
from utils import ListNode
class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
prev = None
while head:
head.next, prev, head = prev, head, head.next
return prev
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Recursively
solution_recursive.py1
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| from typing import Optional
from utils import ListNode
class Solution:
def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
if head is None:
return None
return self._helper(head)
def _helper(self, head: ListNode) -> ListNode:
if (next := head.next) is None:
return head
first = self._helper(next)
next.next = head
head.next = None
return first
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