19.removeNthNodeFromEndofList.md

题目地址

https://leetcode.com/problems/remove-nth-node-from-end-of-list/description

题目描述

Given a linked list, remove the n-th node from the end of list and return its head.

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. Note:

Given n will always be valid.

Follow up:

Could you do this in one pass?

前置知识

• 链表
• 双指针

公司

• 阿里
• 百度
• 腾讯
• 字节

思路

双指针，指针 A 先移动 n 次， 指针 B 再开始移动。当 A 到达 null 的时候， 指针 b 的位置正好是倒数 n

我们可以设想假设设定了双指针 p 和 q 的话，当 q 指向末尾的 NULL，p 与 q 之间相隔的元素个数为 n 时，那么删除掉 p 的下一个指针就完成了要求。

设置虚拟节点 dummyHead 指向 head

设定双指针 p 和 q，初始都指向虚拟节点 dummyHead

移动 q，直到 p 与 q 之间相隔的元素个数为 n

同时移动 p 与 q，直到 q 指向的为 NULL

将 p 的下一个节点指向下下个节点

(图片来自： https://github.com/MisterBooo/LeetCodeAnimation)

关键点解析

1. 链表这种数据结构的特点和使用

2. 使用双指针

3. 使用一个 dummyHead 简化操作

代码

Support: JS, Java

• Javascript Implementation

/*
 * @lc app=leetcode id=19 lang=javascript
 *
 * [19] Remove Nth Node From End of List
 *
 * https://leetcode.com/problems/remove-nth-node-from-end-of-list/description/
 *
 * algorithms
 * Medium (34.03%)
 * Total Accepted:    360.1K
 * Total Submissions: 1.1M
 * Testcase Example:  '[1,2,3,4,5]\n2'
 *
 * Given a linked list, remove the n-th node from the end of list and return
 * its head.
 *
 * 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.
 *
 *
 * Note:
 *
 * Given n will always be valid.
 *
 * Follow up:
 *
 * Could you do this in one pass?
 *
 */
/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} n
 * @return {ListNode}
 */
var removeNthFromEnd = function (head, n) {
  let i = -1;
  const noop = {
    next: null,
  };

  const dummyHead = new ListNode(); // 增加一个dummyHead 简化操作
  dummyHead.next = head;

  let currentP1 = dummyHead;
  let currentP2 = dummyHead;

  while (currentP1) {
    if (i === n) {
      currentP2 = currentP2.next;
    }

    if (i !== n) {
      i++;
    }

    currentP1 = currentP1.next;
  }

  currentP2.next = ((currentP2 || noop).next || noop).next;

  return dummyHead.next;
};

• Java Code

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        TreeNode dummy = new TreeNode(0);
        dummy.next = head;
        TreeNode first = dummy;
        TreeNode second = dummy;

        if (int i=0; i<=n; i++) {
            first = first.next;
        }

        while (first != null) {
            first = first.next;
            second = second.next;
        }

        second.next = second.next.next;

        return dummy.next;
    }
}