【问题描述】
Reverse a singly linked list.
【解答】
class Solution {public:ListNode* reverseList(ListNode* head) {if(head==NULL || head->next==NULL)return head;ListNode* pre=NULL;ListNode* cur=head;while(cur!=NULL){ListNode* nt=cur->next;cur->next=pre;pre=cur;cur=nt;}return pre;}};
这道题不会，考虑了好久，绕不过来~看了答案写的，最初返回的时候还写成返回cur，仔细观察了一下，应该是返回pre。
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我会带着你远行
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