LeetCode31

206.反转链表

给你单链表的头节点 head ，请你反转链表，并返回反转后的链表。

示例 1：

输入：head = [1,2,3,4,5]
输出：[5,4,3,2,1]

示例 2：

输入：head = [1,2]
输出：[2,1]

示例 3：

输入：head = []
输出：[]

提示：

• 链表中节点的数目范围是 [0, 5000]
• -5000 <= Node.val <= 5000

# 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 head is None:return Noneelif head.next is None:return headelse:dummy = ListNode()p = headwhile p.next is not None:p = p.nextdummy.next = pwhile 1:p = headwhile p.next.next is not None:p = p.nextp.next.next = pp.next = Noneif head.next is None:return dummy.next

# 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 head is None:return Noneelif head.next is None:return headelse:q = Nonep = headwhile p is not None:j = p.nextp.next = qq = pp = jreturn q

总结

从后往前，不能遍历到最后一个，只能遍历到倒数第二个。

从前往后，需要三个指针，多一个存储下一个节点。