题解 | #删除链表的倒数第n个节点#

利用dummynode来解决头节点变化的问题；

fast指针向前进n步，拉开距离；

fast指针到达null，slow即在倒数第n个；

这道题给的n都能找到valid的节点，如果找不到倒数nth node，更改部分代码为：

       for (int i = 0; i < n; i++) {
            if(fast != null) fast = fast.next;
        }
        if (fast == null) return head;

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 *   public ListNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head ListNode类 
     * @param n int整型 
     * @return ListNode类
     */
    public ListNode removeNthFromEnd (ListNode head, int n) {
        // write code here
    
        ListNode dummy = new ListNode (-1);
        dummy.next = head;
        ListNode slow = dummy;
        ListNode fast = dummy;
        for (int i = 0; i < n; i++) {
            fast = fast.next;
        }
        while (fast.next != null) {
            fast = fast.next;
            slow = slow.next;
        }
        ListNode prev = slow;
        ListNode toBeDeleted = slow.next;
        ListNode next = toBeDeleted.next;
        prev.next = next;
        toBeDeleted.next = null;
        return dummy.next;
    }
}