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,
:分别讨论三种情况：
1、head指向为空；
2、删除的元素为第一个元素；
3、一般情况；

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

/** * 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) {        if(head == NULL)           return NULL;        ListNode *p = head;             int A = 1;        while (p->next != NULL){            p = p->next;            A = A + 1;        }        int num = A - n;        if( num == 0){            head = head->next;            return head;        }        ListNode *q = head;        for(int i = 1; i < num; i++)            q = q->next;        q->next = q->next->next;          return head;    }};