面试经典-删除链表的倒数第N个节点

删除链表的倒数第N个节点

题目

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]

提示：

• 链表中结点的数目为 sz
• 1 <= sz <= 30
• 0 <= Node.val <= 100
• 1 <= n <= sz

进阶：你能尝试使用一趟扫描实现吗？

题解

此题有点难，适合二刷

import java.util.ArrayList;
import java.util.List;

public class RemoveNthFromEnd {

    public static void main(String[] args) {
        ListNode head = new ListNode(1);
        head.next = new ListNode(2);
        head.next.next = new ListNode(3);
        head.next.next.next = new ListNode(4);
        head.next.next.next.next = new ListNode(5);
        ListNode result = removeNthFromEnd(head, 2);
        while (result != null) {
            System.out.println(result.val);
            result = result.next;
        }
    }

    public static ListNode removeNthFromEnd1(ListNode head, int n) {
        if (head == null) {
            return null;
        }
        List<Integer> list = new ArrayList<>();
        while (head.next != null) {
            list.add(head.val);
            head = head.next;
        }
        list.add(head.val);
        list.remove(list.size() - n);
        ListNode result = null;
        ListNode item = null;
        for (Integer integer : list) {
            if (result == null) {
                result = new ListNode(integer);
                item = result;
            }else{
                item.next = new ListNode(integer);
                item = item.next;
            }
        }
        return result;
    }

    public static ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode(0, head);
        int length = getLength(head);
        ListNode cur = dummy;
        for (int i = 1; i < length - n + 1; ++i) {
            cur = cur.next;
        }
        cur.next = cur.next.next;
        return dummy.next;
    }

    public static int getLength(ListNode head) {
        int length = 0;
        while (head != null) {
            ++length;
            head = head.next;
        }
        return length;
    }
}