2024/11/26【链表】LeetCode 206 反转链表

206. 反转链表 - 力扣（LeetCode）

代码随想录

解法1：双指针法

解法2：递归法（代码如下所示）

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverse(self, pre, cur):
        if cur is None:
            return pre
            
        temp = cur.next
        cur.next = pre
        return self.reverse(cur, temp)

    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        return self.reverse(None, head)