Linked List Random Node

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