题解 | #删除链表的倒数第n个节点#

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 * }
 */

public class Solution {
    /**
     * 
     * @param head ListNode类 
     * @param n int整型 
     * @return ListNode类
     */
    public ListNode removeNthFromEnd (ListNode head, int n) {
        // write code here
        ListNode tmp = head;
        ListNode tmp2 = tmp;
        int num = 0;
        while(head != null){
            num++;
            head = head.next;
        }
        if(num == 0){
            return null;
        }
        if(n == num){
            return tmp.next;
        }
        int k = 0;

        while(tmp != null){
            if(k == num - n - 1){
                tmp.next = tmp.next.next;
                return tmp2;
            }else{
                k++;
                tmp = tmp.next;

            }
        }

        return tmp2;
    }
}

两次遍历，肯定是能解决问题的。但是最优解应该是双指针。p和q，p和q相距n,当q到达末尾时p就是倒数第n个节点。