LeetCode：Reverse Linked List

问题描述：Reverse a singly linked list.顾名思义，就是将单链表反转，这其实一道很简单的算法，我在前面的博客也写过解法，以前是刚看过《剑指offer》，所以对那上面的解法印象较深。

解法一：

参考代码如下：

结构体定义：

struct NodeList
{
int value;
struct NodeList  next;
};
函数实现：
NodeList *ReverseList(NodeList * pHead)
{
NodeList * pNode = pHead;
NodeList * pPrev = NULL;
NodeList * pReverseHead = NULL;
while(pNode != NULL)
{
NodeList *pNext = pNode->next;
if(pNext == NULL)
{
pReverseHead = pNode;
}
pNode->next = pPrev;
pPrev = pNode;
pNode = pNext;
}
return pReverseHead;
}

今天又刷到这道题目，这次是用头插法实现。

解法二：

/**
* 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* pPre = head;
ListNode* pCurrent = head -> next;
pPre -> next = NULL;
while(pCurrent)
{
ListNode* pNext = pCurrent -> next;
if(pCurrent -> next == NULL)
{
pCurrent -> next = pPre;
return pCurrent;
}
pCurrent -> next = pPre;
head = pCurrent;
pPre = pCurrent;
pCurrent = pNext;
}
return head;
}
};