Leetcode: 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();

Solution 1: Reservior sampling: (wiki introduction)

Reservoir sampling is a family of randomized algorithms for randomly choosing a sample of k items from a list S containing n items, where n is either a very large or unknown number. Typically n is large enough that the list doesn't fit into main memory.

example: size = 1

Suppose we see a sequence of items, one at a time. We want to keep a single item in memory, and we want it to be selected at random from the sequence. If we know the total number of items (n), then the solution is easy: select an index i between 1 and n with equal probability, and keep the i-th element. The problem is that we do not always know n in advance. A possible solution is the following:

• Keep the first item in memory.
• When the i-th item arrives (for i>1):
  • with probability 1/i, keep the new item (discard the old one)
  • with probability 1-1/i, keep the old item (ignore the new one)

So:

• when there is only one item, it is kept with probability 1;
• when there are 2 items, each of them is kept with probability 1/2;
• when there are 3 items, the third item is kept with probability 1/3, and each of the previous 2 items is also kept with probability (1/2)(1-1/3) = (1/2)(2/3) = 1/3;
• by induction, it is easy to prove that when there are n items, each item is kept with probability 1/n.

 

This problem is size=1

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    ListNode start;

    /** @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.start = head;
    }
    
    /** Returns a random node's value. */
    public int getRandom() {
        Random random = new Random();
        ListNode cur = start;
        int val = start.val;
        
        for (int i=1; cur!=null; i++) {
            if (random.nextInt(i) == 0) {
                val = cur.val;
            }
            cur = cur.next;
        }
        return val;
    }
}

/**
 * Your Solution object will be instantiated and called as such:
 * Solution obj = new Solution(head);
 * int param_1 = obj.getRandom();
 */