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206. 反转链表 - 力扣（LeetCode）https://leetcode.cn/problems/reverse-linked-list/solutions/551596/fan-zhuan-lian-biao-by-leetcode-solution-d1k2/?envType=study-plan-v2&envId=top-100-liked

常规法

记录前一个指针，当前指针，后一个指针，遍历链表，改变指针的指向。

//c++
/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* reverseList(ListNode* head) {ListNode *pre=nullptr;while(head){ListNode *nex=head->next;head->next=pre;pre=head;head=nex;}return pre;}
};#python
# 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]:pre=Nonewhile head:nex=head.nexthead.next=prepre=headhead=nexreturn pre  

递归法 

对于每一个元素的下一个的下一个要指向自己，自己再指向空，递归进去得到头指针。

//c++
/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* reverseList(ListNode* head) {if(!head||!head->next)  return head;ListNode * ne=reverseList(head->next);head->next->next=head;head->next=nullptr;return ne;}
};#python
# 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 or not head.next:return headnew=self.reverseList(head.next)head.next.next=head;head.next=None;return new