Reservoir Sampling

382. Linked List Random Node
LeetCode上的一道题

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?

思路

已知链表长度n时，可以使用

rand()%n  //

但此题目提示，链表长度未知，此时需要用到一些概率论的知识。

1. 每个元素被抽取概率均为1/n时，采样等概
2. 考虑可覆盖的采样，即后一个采样可以覆盖前一个采样。
3. 第k个元素，只考虑前k，对应概率为1/k
4. 假设第i个元素，被采中。对应事件为第i次采中，之后所有采样均未采中，计算如下
   
   1i⋅(1−1i+1)…(1−1n)=1n
   
   即第i个元素被采中的概率为1/n

  int getRandom() {    ListNode *tem = p;    int val = p->val;    for (int i = 1; tem; i++) {        if (rand() % i == 0)            val = tem->val;        tem = tem->next;    }    return val;          }