初级算法 : 反转链表 ----- 递归法
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
进阶:链表可以选用迭代或递归方式完成反转。你能否用两种方法解决这道题?
作者:力扣 (LeetCode)
链接:https://leetcode.cn/leetbook/read/top-interview-questions-easy/xnnhm6/
来源:力扣(LeetCode)
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。
递归法:
/** * 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 == NULL || head->next == NULL){ return head; } ListNode* next = head->next; ListNode* reverse = reverseList(next);//递归 next->next = head; head->next = NULL; return reverse; } };
迭代 双链表法:
/**
* 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* newHead = NULL;
while(head != NULL){//迭代
ListNode* temp = head->next;
head->next = newHead;
newHead = head;
head = temp;
}
return newHead;
}
};
![](https://img2022.cnblogs.com/blog/2986763/202210/2986763-20221027165229070-1505352389.png)
hello my world
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