[leetcode_19] Remove Nth Node From End of List

Given a singly linked list, remove the nth node from the end. Try to do it in one pass.
Use two pointers with a gap of exactly n between them. When the second pointer reaches NULL, remove the node at the first pointer.
Here is the code:

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class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        ListNode *before = NULL;
        ListNode *p1 = head;
        ListNode *p2 = head;
        if(n <= 0) return head;
        for(int i = 0; i < n; i++) {
            p2 = p2->next;
        }
        while(p2 != NULL) {
            p2 = p2->next;
            before = p1;
            p1 = p1->next;
        }
        if(before == NULL) {
            return head->next;
        } else {
            before->next = p1->next;
            return head;
        }
    }
};