19. 删除链表的倒数第 N 个结点

给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。

示例 1：

输入：head = [1,2,3,4,5], n = 2
输出：[1,2,3,5]
示例 2：

输入：head = [1], n = 1
输出：[]
示例 3：

输入：head = [1,2], n = 1
输出：[1]

最容易理解的：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        if (head == null) return null;
        // 先算一下一共有多少个节点
        int count = 0;
        ListNode p = head;
        while (p != null) {
            count ++;
            p = p.next;
        }

        ListNode result = new ListNode(0, head);
        // 算一下要移动多少步
        int needMoveNum = count - n;
        p = result;
        while (needMoveNum > 0) {
            p = p.next;
            needMoveNum --;
        }

        // 操作
        p.next = p.next.next;

        // 结束
        return result.next;
    }
}

栈

class Solution {
   public ListNode removeNthFromEnd(ListNode head, int n) {
		Deque<ListNode> stack = new LinkedList<ListNode>();
		ListNode result = new ListNode(0, head);
		ListNode p = result;
		while (p != null) {
			stack.push(p);
			p = p.next;
		}

		for (int i = 0; i < n; i++) {
			stack.pop();
		}
		ListNode oper = stack.peek();
		oper.next = oper.next.next;
		return result.next;
    }
}