C++版 - 剑指offer 面试题16:反转链表(Leetcode 206: Reverse Linked List) 题解

面试题16:反转链表

提交网址: http://www.nowcoder.com/practice/75e878df47f24fdc9dc3e400ec6058ca?tpId=13&tqId=11168

或 https://leetcode.com/problems/reverse-linked-list/

Total Accepted: 101523  Total Submissions: 258623  Difficulty: Easy

Reverse a singly linked list.

Hint:

A linked list can be reversed either iteratively or recursively. Could you implement both?

  • 参与人数:5517  时间限制:1秒  空间限制:32768K
  • 本题知识点:链表

分析:

使用头插法,并每次将newhead获取到最前端临时结点(整体赋值)...

有空了,再来用递归思想实现一次...

AC代码:

class Solution {
public:
ListNode* ReverseList(ListNode* pHead) { ListNode *p;
ListNode *newhead=NULL;
p=pHead; if(pHead==NULL || pHead->next==NULL) return pHead; while(p!=NULL)
{
ListNode *temp=p;
p=p->next; temp->next=newhead; // 挂接上
newhead=temp; // 将新插入的节点整体复制给头指针结点
}
return newhead;
}
};
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