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;
	}
};

Remove Nth Node From End of List

原文：http://blog.csdn.net/sxhlovehmm/article/details/44783385