[LeetCode]Remove Nth Node From End of List

Given a linked list, remove the n^th 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.

思考：设两个指针一前一后间隔n-1个元素。分两种情况。

/**
 * 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) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        ListNode *p,*q,*r;
		p=head;
		r=head;
		while(n--)
			p=p->next;
		if(p==NULL) 
		{
			head=r->next;
			delete r;
		}
		else
		{
			while(p->next)
			{
				r=r->next;
				p=p->next;
			}
			q=r->next;
			r->next=q->next;
			delete q;
		}
		return head;
    }
};