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.

Analyse: 如果链表为空，返回NULL；如果n等于链表的长度，即删除head；否则，先找到该删除位置的前一位置，将该位置的next指向被删除位置的next，并释放被删除指针的空间。第一次没有WA或TE就AC了，mark一下~~ :)

/**
 * 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 *current = head;
        int length = 0;
        while(current){
            length++;
            current = current->next;
        }
        
        if(length == 1) return NULL;
        if(length == n){
            ListNode *temp = head;
            head = head->next;
            delete temp;
            return head;
        }
        
        current = head;
        for(int i = 1; i < length - n; i++) current = current->next;
        
        ListNode *temp = current->next;
        current->next = current->next->next;
        delete temp;
        
        return head;
    }
};