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**.

Solution:

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