LeetCode 19. Remove Nth Node From End of List

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

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.

删除链表的倒数第n个节点
声明一个slow指针和一个fast指针，fast指针比slow指针领先（n+1）个节点，当fast指针指向NULL时，删除slow指向的节点即可

class Solution {public:    ListNode* removeNthFromEnd(ListNode* head, int n) {        ListNode *slow=head,*fast=head;        while(n){            fast=fast->next;            n--;        }        if(fast==NULL)            return slow->next;        fast=fast->next;        while(fast){            slow=slow->next;            fast=fast->next;        }        slow->next=slow->next->next;        return head;    }};