题目链接:https://leetcode.com/problems/reverse-linked-list/
方法一:迭代反转
https://blog.csdn.net/qq_17550379/article/details/80647926讲的很清楚
/** * 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; ListNode* lat; while(cur){ lat = cur -> next; cur -> next = pre; pre = cur; cur = lat; } return pre; } };
方法二:递归反转
解决递归问题从最简单的case入手。
(1)例如,对于链表1->2->3->4->5,
reverseList(head)返回的是5->4->3->2->1;
reverseList(head->next)返回的是5->4->3->2;(因为head->next所指代的链表是2->3->4->5->NULL)
以此类推。
(2)对于reverseList(3)这个情况,此时head为3,head->next为4,此时链表情况如下:
1->2->3->4<-5
head->next->next=head这一操作后,链表变为:
然后,head->next=NULL,即
1->2->3<-4<-5
/** * 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) { if(head == NULL || head -> next == NULL) return head; ListNode *ans = reverseList(head -> next); head -> next -> next = head; head -> next = NULL; return ans; } };
此外,提供一个错误写法:
/** * 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) { if(head == NULL || head -> next == NULL) return head; ListNode *ans = reverseList(head -> next); ans -> next = new ListNode(head -> val); return ans; } };
这种写法的错误之处在于:
以reverseList(3)为例,虽然ListNode *ans = reverseList(head -> next);返回的是5->4这个链表,不过ans指的是这个链表的第一个结点5,然后ans->next=new ListNode(head->val)后,其实在当前这一层递归中最终return的是5->3这个链表,并不是返回想象中的5->4->3这个链表。
参考:https://blog.csdn.net/qq_17550379/article/details/80647926