[LeetCode]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.

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;
}
};