Remove Nth Node From End of List

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.

/**
* 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) {
if(head==NULL)
return NULL;
ListNode *pre=NULL,*p=head,*q=head;
for(int i=0;i<n-1;i++){
q=q->next;
}//退出循环后，如果q->next为空，则说明删除的是首结点
while(q->next){
pre=p;
p=p->next;
q=q->next;
}
if(pre==NULL){//pre==NULL说明删除的是首结点
head=p->next;
delete p;
}else{
pre->next=p->next;
delete p;
}
return head;
}
};