LeetCode-019：Remove Nth Node From End of List

题目：

Given a linked list, remove the n-th node from the end of list and return its head.

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.

Follow up:

Could you do this in one pass?

题意：

一遍遍历删除链表中倒数第n个节点

思路：

重新建立链表，设立左右两个指针，先让右指针向前走n步；然后左右两个指针向前遍历直到右边的指针到达尾结点，然后删除左指针的下一个节点（即倒数第n个节点）~Runtime: 4 ms, faster than 100.00% of C++ online submissions for Remove Nth Node From End of List.~第一次100%，即使是水题~

Code：

/**
 * 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) {
        ListNode *t=new ListNode(0),*l=t,*r=t;
        t->next=head;
        while(n--) r=r->next;
        while(r->next){
            l=l->next;
            r=r->next;
        }
        l->next=l->next->next;
        return t->next;
    }
};