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.

思路：先从头遍历等到长度length，在从头遍历，根据当前结点次序与长度length和n的关系确定要删除的结点。

/**
 * 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 *p=head,*q;
        int length=0;
        while(p!=NULL)
        {
            length++;
            p=p->next;
        }
        p=head;
        head=q;
        q->next=p;
        while(p!=NULL)
        {
            if(length==n)
            {
               q->next=p->next;
               break;
            }
            else
            {
                q=p;
                p=p->next;
                length--;
            }
        }
        return head->next;
    }
};