方法一 递归
前后指正的转换，需要一个临时指针来存后一个节点，最后实现当前指针的后移。
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* reverseList(ListNode* head) { ListNode* pre = NULL; ListNode* cur = head; while(cur != NULL){ ListNode* tmp = cur -> next; //tmp = old cur -> next cur -> next = pre; //old pre pre = cur; //new pre = old pre cur = tmp; // new cur = tmp(old cur -> next) } return pre; } };
方法二 递归
假设链表为1->2->3->4->5，层层递归
1、当递归到最后reverseList(5)(head=head->next=4->next=5)时,head->next=NULL返回head(即5，p指向5),这层递归结束。
2、reverseList(5)(head->next=4->next=5)后的代码是head->next->next = head(5->next=4即5->4);head->next = NULL(4->NULL);return p(返回5->4->NULL);
3、再看reverseList(4)(head->next=3->next=4)后中后head->next->next = head(4->3);head->next = NULL(3->NULL);return p(返回5->4->3->NULL);
4、reverseList(2), reverseList(1)依次类推，p最终为5->4->3->2->1->NULL。
作者：24shi-01fen-_00_01
链接：https://leetcode-cn.com/problems/reverse-linked-list/solution/206fan-zhuan-lie-biao-die-dai-di-gui-by-24shi-01fe/
class Solution { public: ListNode* reverseList(ListNode* head) { if (head == NULL || head->next == NULL) return head; ListNode* p = reverseList(head->next); head->next->next = head; head->next = NULL; return p; } };
原文：https://www.cnblogs.com/BillowJ/p/12679541.html