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();

A probability problem. The easiest solution is to traverse the whole list, find the count (number of nodes in the list) and randomly pick one node with 1 / count probability. As the question mentioned, if the list is too long, it will be too hard to get the length. One approach is to keep counting and choose the position at 1 / count. Traverse through the whole list and return the last number. The intuition is, for any number x, the probability is : 1 / x * (x / (x + 1))... (n - 1)/n = 1 / n.

public class Solution {
    
    ListNode head;
    Random random;

    /** @param head The linked list's head.
        Note that the head is guaranteed to be not null, so it contains at least one node. */
    public Solution(ListNode head) {
        this.head = head;
        random = new Random();
    }
    
    /** Returns a random node's value. */
    public int getRandom() {
        ListNode curr = head;
        int toReturn = 0;
        for (int cnt = 1; curr != null; cnt++, curr = curr.next) {
            if (random.nextInt(cnt) == 0) {
                toReturn = curr.val;
            }
        }
        return toReturn;
    }
}