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.

思路：保持两个指针的距离是n-1，然后当尾指针遍历到最后一个的时候，头指针就指向了第n个，然后处理一下就行了

/**
 * 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 *tmp, *s, *e;
        tmp = s = e = head;

        while (n > 1) {
            e = e->next;
            n--;
        }

        while (e->next != NULL) {
            tmp = s;
            s = s->next;
            e = e->next;
        }

        if (tmp == s) return head->next;
        else tmp->next = s->next;

        return head;
    }
};

LeetCode Remove Nth Node From End of List