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();

本题可以用reservoir sampling来解决不明list长度的情况下平均概率选择元素的问题。

假设在[x_1,...,x_n]只选一个元素，要求每个元素被选中的概率都是1/n，但是n未知。 其中 random.randint(0, cnt) == 0: 的概率是1/(cnt+1)。reservoir sampling的证明可以使用归纳法(induction)。

 class Solution(object):

     def __init__(self, head):

         """

         @param head The linked list's head.

         Note that the head is guaranteed to be not null, so it contains at least one node.

         :type head: ListNode

         """

         self.head = head

     def getRandom(self):

         """

         Returns a random node's value.

         :rtype: int

         """

         cnt = 0

         head = self.head

         while head:

             if random.randint(0, cnt) == 0:

                 ans = head.val

             head = head.next

             cnt += 1

         return ans

本题的一个推广是如何在[x_1,...,x_n]中选出k个元素。并且每个x_i被选中的概率都一样，而且n未知。

1. if i <= k, T_i = T_{i-1}\cup x_i

2. else with p = k/i replace one element in T_{i-1} with x_i with p=1/k。

证明同样可以用归纳法。