标签：
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();
解法一：暴力解法
import java.util.Random; public class Solution { private ListNode head; private int size; Random randGen = new Random(); /** @param head The linked list‘s head. Note that the head is guanranteed to be not null, so it contains at least one node. */ public Solution(ListNode head) { this.head = head; ListNode current = head; while (current != null) { size++; current = current.next; } this.size = size; } /** Returns a random node‘s value. */ public int getRandom() { int pos = randGen.nextInt(size); ListNode current = head; for (int i =0; i < pos; i++) current = current.next; return current.val; } }
解法二：蓄水池抽样
在遍历到第i个数时设置选取这个数的概率为1/i, 然后来证明一下每个数被选到的概率:
对于第一个数其被选择的概率为1/1*(1-1/2)*(1-1/3)*(1-1/4)*...*(1-1/n) = 1/n, 其中(1-1/n)的意思是不选择n的概率,
也就是选择1的概率乘以不选择其他数的概率. 对于任意一个数i来说, 其被选择的概率为1/i*(1-1/(i+1))*...*(1-1/n),
所以在每一个数的时候我们只要按照随机一个是否是i的倍数即可决定是否取当前数即可.
详细证明和讲解：
http://blog.csdn.net/yeqiuzs/article/details/52169369
所以当遍历到第cnt个数的时候，保证选取这个数的概率为1/cnt，即可最后保证所有数被抽取的概率相等，皆为1/n
import java.util.Random; public class Solution { private ListNode head; private Random random; /** @param head The linked list‘s head. Note that the head is guanranteed to be not null, so it contains at least one node. */ public Solution(ListNode head) { this.head = head; this.random = new Random(); } /** Returns a random node‘s value. */ public int getRandom() { int ans = 0; ListNode p = head; for (int cnt = 1; p != null; cnt++, p = p.next) if (random.nextInt(cnt) == 0) ans = p.val;//nextInt(cnt)产生0-cnt随机数，其中等于0的概率正好为1/cnt return ans; } }
reference:
https://www.hrwhisper.me/leetcode-linked-list-random-node/
http://blog.csdn.net/yeqiuzs/article/details/52169369
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原文地址：http://www.cnblogs.com/hygeia/p/5808928.html
