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();

public class Solution {
    /**
     * Start... When we read the first node head, if the stream ListNode stops here,
     * we can just return the head.val. The possibility is 1/1.
     * 
     * When we read the second node, we can decide if we replace the result r or
     * not. The possibility is 1/2. So we just generate a random number between 0
     * and 1, and check if it is equal to 1. If it is 1, replace r as the value of
     * the current node, otherwise we don't touch r, so its value is still the value
     * of head.
     * 
     * When we read the third node, now the result r is one of value in the head or
     * second node. We just decide if we replace the value of r as the value of
     * current node(third node). The possibility of replacing it is 1/3, namely the
     * possibility of we don't touch r is 2/3. So we just generate a random number
     * between 0 ~ 2, and if the result is 2 we replace r.
     * 
     * We can continue to do like this until the end of stream ListNode.
     */
    ListNode head = null;
    Random randomGenerator = null;

    public Solution(ListNode head) {
        this.head = head;
        this.randomGenerator = new Random();

    }

    /** Returns a random node's value. */
    public int getRandom() {
        ListNode result = null;
        ListNode current = head;

        for (int n = 1; current != null; n++) {
            if (randomGenerator.nextInt(n) == 0) {
                result = current;
            }
            current = current.next;
        }
        return result.val;
    }
}