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.

解法：从后面往前数的第n个
pointerA先从前开始数n个，然后另一个pointerB从开头数，当A到达最后的时候,b就到了合正确的地方

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode record=new ListNode(0);
        record.next=head;
        ListNode node=record;
        for(int i=0;i<n;i++){
            if(head==null){
                return null;
            }
            head=head.next;
        }
        while(head!=null){
            head=head.next;
            node=node.next;
            
        }
        node.next=node.next.next;
        return record.next;
    }
}