LeetCode OJ 之 Remove Nth Node From End of List ( 删除链表的从尾部数第n个结点 ）

题目：

Given a linked list, remove the nth node from the end of list and return its head.

删除链表的从尾部数第n个结点，返回头结点

For example,

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:

Given n will always be valid.

Try to do this in one pass.
代码：
方法一：

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *removeNthFromEnd(ListNode *head, int n)
{
//常规思路：根据链表总长度算出倒数第n个结点位于正数第几个结点（len+1-n），再正序遍历，然后删除。
int len=1;//链表长度
ListNode *p=head;
if(head == NULL || n == 0)
return NULL;
while(p->next != NULL)//计算链表长度
{
len++;
p = p->next;
}
if(len < n)
return NULL;
int num = len + 1 - n;//正着数时结点的位置
ListNode *q=head;//定义q用来记录p结点的前一个结点，方便删除
p=q->next;
if(num == 1)
{
head = p;
return head;
}
int i=1;
while(i < num-1)//查找结点，p为要删除的结点
{
q=p;
p=p->next;
i++;
}
q->next = p->next;//删除p
delete p;
return head;
}
};

方法二：

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *removeNthFromEnd(ListNode *head, int n)
{
ListNode dummy{-1, head};//定义虚拟头结点
ListNode *p = &dummy, *q = &dummy;
for (int i = 0; i < n; i++) //q先走n步
q = q->next;
while(q->next) //q到达尾结点，p、q都走了len-n步，那么p就到达要删除结点的前一个结点
{
p = p->next;
q = q->next;
}
ListNode *tmp = p->next;//tmp为要删除的结点
p->next = p->next->next;
delete tmp;
return dummy.next;
}
}
};