206 - Reverse Linked List

Reverse a singly linked list.

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思路分析：
此题较容易。
我们可以设置一个新的指针newHead。初始化为NULL。
此时，迭代原始的链表。每一次把原始的链表的节点从头插入newHead中。
假定初始链表为：1->2->3->4->5
1. newhead->1->null

2. newhead->2->1->null

...

5 newhead->5->4->3->2->1->null

Then just return newhead->next.

#include "stdafx.h"
#include <iostream>

using namespace std;

struct ListNode 
{
	int val;
	ListNode *next;
	ListNode(int x) : val(x), next(NULL) {}
};

class Solution_206_ReverseLinkedList
{
public:
	ListNode* reverseList(ListNode* head) 
	{
		ListNode *newHead = new ListNode(0);

		while (head)
		{
			ListNode *tmp = head->next;
			newHead->next = head;
			head = head->next;
			newHead->next->next = tmp;
		}
		return newHead->next;
	}
};