2024/11/26【链表】LeetCode 206 反转链表
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)