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.

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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(head == NULL || n < 1)return head;
    	ListNode* p1 = head;
		ListNode* p2 = head;
		while(n>0 && p1 != NULL) {p1=p1->next;n--;}
		if(p1 == NULL){
			if(n==0)
			{
                head = p2->next;
                delete p2;
                return head;
			}
			else return head;
		}
		while(p1->next != NULL)
		{p1=p1->next;p2=p2->next;}
		p1=p2->next;
		p2->next = p1->next;
		delete p1;
		return head;
    }
};