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?


// Init a singly linked list [1,2,3].
ListNode head = new ListNode(1); = new ListNode(2); = 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.


Use Reservoir Sampling.


Choose k entries from n numbers. Make sure each number is selected with the probability of k/n.

Basic idea:

  1. Choose 1, 2, 3, ..., k first and put them into the reservoir.
  2. For k+1, pick it with a probability of k/(k+1) (this can be done via Random.nextInt(k+1) < k), and randomly replace a number in the reservoir.
  3. For k+i, pick it with a probability of k/(k+i), and randomly replace a number in the reservoir.
  4. Repeat until k+i reaches n.


  1. For k+i, the probability that it is selected and will replace a number in the reservoir is k/(k+i)
  2. For a number in the reservoir before (let’s say x), the probability that it keeps staying in the reservoir is
P(x was in the reservoir last time) × P(X is not replaced by k+i)
= P(X was in the reservoir last time) × (1 - P(k+i is selected and replaces X))
= k/(k+i-1) × (1 - k/(k+i) × 1/k)
= k/(k+i)
  1. When k+i reaches n, the probability of each number staying in the reservoir is k/n.

An implementation of Reservoir Sampling is as follows:

int[] reservoir = new int[k];
Random random = new Random(System.currentTimeMillis());
for(int i = 0; i < k; i++) reservoir[i] = input[i];
for(int j = k; k < n; k++) {
    int rnd = random.nextInt(j);
    if(rnd < k)
        reservoir[rnd] = input[j];
return reservoir;

The solution of this question is as follows:

 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
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;
        this.random = new Random(System.currentTimeMillis());

    /** Returns a random node's value. */
    public int getRandom() {
        if(head == null) return -1;
        ListNode res = head, node = head;
        int i = 1;
        while( != null) {
            node =;
            if(random.nextInt(++i) == 0)
                res = node;
        return res.val;

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