Linked List Random Node

本文介绍了一种无需预先知道链表长度就能高效获取随机节点的方法,采用水库抽样算法实现每个节点被选中的概率相等。

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Given a singly linked list, return a random node’s value from the linked list. Each node must have the same probability of being chosen.

Follow up:
What if the linked list is extremely large and its length is unknown to you? Could you solve this efficiently without using extra space?

Example:

// Init a singly linked list [1,2,3].
ListNode head = new ListNode(1);
head.next = new ListNode(2);
head.next.next = new ListNode(3);
Solution solution = new Solution(head);

// getRandom() should return either 1, 2, or 3 randomly. Each element should have equal probability of returning.
solution.getRandom();

解:
虽然可以初始化的时候遍历一遍列表求的它的长度,但本题的意思是在不提前求得列表长度的时候,列表以顺序流的方式访问,如何得到一个随机的节点。

其实这是一个reservoir sampling的问题

https://en.wikipedia.org/wiki/Reservoir_sampling
PROBLEM:

Choose k entries from n numbers. Make sure each number is selected with the probability of k/n
BASIC IDEA:

Choose 1, 2, 3, …, k first and put them into the reservoir.
For k+1, pick it with a probability of k/(k+1), and randomly replace a number in the reservoir.
For k+i, pick it with a probability of k/(k+i), and randomly replace a number in the reservoir.
Repeat until k+i reaches n
PROOF:

For k+i, the probability that it is selected and will replace a number in the reservoir is k/(k+i)
For a number in the reservoir before (let’s say X), the probability that it keeps staying in the reservoir is
P(X was in the reservoir last time) × P(X is not replaced by k+i)
P(X was in the reservoir last time) × (1 - P(k+i is selected and replaces X))
= k/(k+i-1) × (1 - k/(k+i) × 1/k)
= k/(k+i)
When k+i reaches n, the probability of each number staying in the reservoir is k/n
EXAMPLE

Choose 3 numbers from [111, 222, 333, 444]. Make sure each number is selected with a probability of 3/4
First, choose [111, 222, 333] as the initial reservior
Then choose 444 with a probability of 3/4
For 111, it stays with a probability of
P(444 is not selected) + P(444 is selected but it replaces 222 or 333)
= 1/4 + 3/4*2/3
= 3/4
The same case with 222 and 333
Now all the numbers have the probability of 3/4 to be picked
THIS PROBLEM

This problem is the sp case where k=1

资源下载链接为: https://pan.quark.cn/s/f989b9092fc5 HttpServletRequestWrapper 是 Java Servlet API 中的一个工具类,位于 javax.servlet.http 包中,用于对 HttpServletRequest 对象进行封装,从而在 Web 应用中实现对 HTTP 请求的拦截、修改或增强等功能。通过继承该类并覆盖相关方法,开发者可以轻松地自定义请求处理逻辑,例如修改请求参数、添加请求头、记录日志等。 参数过滤:在请求到达处理器之前,可以对请求参数进行检查或修改,例如去除 URL 编码、过滤敏感信息或进行安全检查。 请求头操作:可以修改或添加请求头,比如设置自定义的 Content-Type 或添加认证信息。 请求属性扩展:在原始请求的基础上添加自定义属性,供后续处理使用。 日志记录:在处理请求前记录请求信息,如 URL、参数、请求头等,便于调试和监控。 跨域支持:通过添加 CORS 相关的响应头,允许来自不同源的请求。 HttpServletRequestWrapper 通过继承 HttpServletRequest 接口并重写其方法来实现功能。开发者可以在重写的方法中添加自定义逻辑,例如在获取参数时进行过滤,或在读取请求体时进行解密。当调用这些方法时,实际上是调用了包装器中的方法,从而实现了对原始请求的修改或增强。 以下是一个简单的示例,展示如何创建一个用于过滤请求参数的包装器: 在 doFilter 方法中,可以使用 CustomRequestWrapper 包装原始请求: 这样,每当调用 getParameterValues 方法时,都会先经过自定义的过滤逻辑。 HttpServletRequestWrapper 是 Java Web 开发中一个强大的工具,它提供了灵活的扩展性,允许开发者
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