Reservoir Sampling-382. 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();

class Solution {
public:
    /** @param head The linked list's head.
        Note that the head is guaranteed to be not null, so it contains at least one node. */
    Solution(ListNode* head):cur(head) {
    }    /** Returns a random node's value. */
    int getRandom() {
        int val=cur->val;
        ListNode *temp=cur;
        for(int i=;temp!=nullptr;temp=temp->next,++i)
        {
            uniform_int_distribution<unsigned> u(,i);
            default_random_engine e(rand());//真正随机的种子
            unsigned int m=u(e);
            if(m<)
            {
                val=temp->val;
            }
        }
        return val;
    }
private:
    ListNode *cur;
};