382. Linked List Random Node (蓄水池抽样，java)

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本文介绍了一种使用蓄水池抽样算法从链表中等概率随机选取节点的方法，并提供了详细的实现步骤及代码示例。

382. Linked List Random Node

Description Submissions Solutions

Total Accepted: 23716
Total Submissions: 51037
Difficulty: Medium
Contributor: LeetCode

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();

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题目链接：https://leetcode.com/problems/linked-list-random-node/#/description

解题思路：题目意思是给定一个链表，随机输出链表中的一个数，要求每个数输出的概率相等。蓄水池抽样可定义为：在不知总个数的情况下，等概率地抽取这组数中的某个数。核心思想是：1、初始答案为第一个数，此时链表的下标指向第一个数，即此时第一个数被选中的概率为1；

2、下标后移一位指向第二个数，用Random函数随机抽取0-1的数，抽取的范围是2，抽中1的概率为1/2，如果抽中1，把答案改为此时下标所指的数，否则不改变答案的值。

3、以此类推，用Random函数抽取的范围不断加1，即Random rd = new Random(i)，抽取范围为i，从0 -（ i－1）中取道 i-1 的概率为1 / i。如果抽中i-1，把答案改为此时下标所指的数，否则不改变答案的值。

4、直到链表为空，得到答案；

第 i 个数被选中的概率为它被选中的概率：1 / i ，乘以后面的数不被选中的概率：[ i / ( i + 1 ) ] * [ ( i + 1 ) / ( i + 2 ) ] *... * [ ( n - 1 ) / ( n + 1 )]

即P（第 i 个数被选中）＝（ 1 / i ）* [ i / ( i + 1 ) ] * [ ( i + 1 ) / ( i + 2 ) ] *... * [ ( n - 1 ) / ( n + 1 )] = 1 / n

代码如下：

（leetcode提交代码）

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {

    static ListNode h = null;
    /** @param head The linked list's head.
        Note that the head is guaranteed to be not null, so it contains at least one node. */
    public Solution(ListNode head) {
        this.h = head; 
    }
    
    /** Returns a random node's value. */
    public int getRandom() {
        Random rd = new Random ();
        int ans = -1;
        int idx = 0;
		ListNode p = h;
        while(p!=null){
        	idx++;
        	int rn = rd.nextInt(idx);
        	if(rn==idx-1){
        		ans = p.val;
        	}
        	p = p.next;
        }
        return ans;
    }
}

/**
 * Your Solution object will be instantiated and called as such:
 * Solution obj = new Solution(head);
 * int param_1 = obj.getRandom();
 */

eclipse 上代码：

import java.util.Random;

public class Main {

	static class ListNode {
		      int val;
		      ListNode next;
		      ListNode(int x) 
		      { val = x; }
		  }
	static ListNode h = null;
    public Main(ListNode head) {
        this.h = head; 
    }
    public static int getRandom() {
        Random rd = new Random ();
        int ans = -1;
        int idx = 0;
		ListNode p = h;
        while(p!=null){
        	idx++;
        	int rn = rd.nextInt(idx);
        	if(rn==idx-1){
        		ans = p.val;
        	}
        	p = p.next;
        }
        return ans;
    }
    public static void main(String[] args){
    	ListNode head = new ListNode(1);
    	head.next = new ListNode(2);
    	head.next.next = new ListNode(3);
    	Main obj = new Main(head);
    	System.out.println(obj.getRandom());
    }
}