链表面试题之单链表逆置

题目：反转一个单链表

链表结构如下：

struct ListNode
{
	int val;
	struct ListNode *next;
};

方法一：

算法思想：用三个指针 n1 , n2 , n3 

起初，n1指向头head，n2指向第二个结点，n3指向第三个结点；先让n1的next指向空，循环判断n2是否为空，让n2头插到n1后面（n2->next = n1），再挪动指针n1，n2，n3。

代码如下： 

struct ListNode* reverseList(struct ListNode* head)
{
	if (head == NULL || head->next == NULL)
		return head;

	struct ListNode* n1, *n2, *n3;
	n1 = head;
	n2 = n1->next;
	n3 = n2->next;

	n1->next = NULL;
	while (n2)
	{
		n2->next = n1;
		n1 = n2;
		n2 = n3;
		if (n3 != NULL)
			n3 = n3->next;
	}
	return n1;
};

方法二：

新创建一个结点newHead = NULL，拿下来结点进行头插，直到最后一个结点

struct ListNode* reverseList(struct ListNode* head)
{
	struct ListNode* cur = head;
	struct ListNode* newHead = NULL;
	while (cur)
	{
		struct ListNode* next = cur->next;
		cur->next = newHead;
		newHead = cur;
		cur = next;
	}
	return newHead;
}