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;        
    }
};