力扣刷题-热题100题-第23题（c++、python）

weixin_445054722025-03-25 19:00

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。

复制代码

//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=None
        while head:
            nex=head.next
            head.next=pre
            pre=head
            head=nex
        return 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。

复制代码

//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 head
        new=self.reverseList(head.next)
        head.next.next=head;
        head.next=None;
        return new