剑指offer——26反转链表

题目描述

输入一个链表,反转链表后,输出新链表的表头。
 
题解:
  每次只反转一个节点,先记住cur->next, 然后pre->cur,即可;
 
 1 class Solution {
 2 public:
 3     ListNode* ReverseList(ListNode* pHead) {
 4         if (pHead == nullptr || pHead->next == nullptr)return pHead;
 5         ListNode  *pre = nullptr, *cur = nullptr, *next = nullptr;
 6         cur = pHead;
 7         while (cur != nullptr)
 8         {
 9             next = cur->next;
10             cur->next = pre;
11             pre = cur;
12             cur = next;
13         }
14         return pre;
15     }
16 };
原文地址:https://www.cnblogs.com/zzw1024/p/11669003.html