`Reservoir Sampling`

Medium

Given a singly linked list, return a random node's value from the linked list. Each node must have thesame probabilityof being chosen.

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].

// getRandom() should return either 1, 2, or 3 randomly. Each element should have equal probability of returning.
solution.getRandom();
``````

## Solution

#### Reservoir Sampling

O(1) space, O(n) time

``````/**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
class Solution {
private Random rand;

Note that the head is guaranteed to be not null, so it contains at least one node. */
rand = new Random();
}

/** Returns a random node's value. */
public int getRandom() {
int result = -1;
int count = 0;

while (p != null) {
count++;
if (rand.nextInt(count) < 1) {
result = p.val;
}
p = p.next;
}
return result;
}
}

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

#### A more generic one for k > 1 reservoir sampling

``````public class Solution {
private Random rand;
Note that the head is guaranteed to be not null, so it contains at least one node. */
this.rand = new Random();
}

/** Returns a random node's value. */
public int getRandom() {
int k = 1;
int res = node.val;
int i = 0;
ArrayList<Integer> reservoir = new ArrayList<Integer>();

while (i < k && node != null) {
node = node.next;
i++;
}
i++; // i == k  =>  i == k+1
while (node != null) {
int j = rand.nextInt(i);
if (j < k) {
reservoir.set(j, node.val);
}
i++;
node = node.next;
}
return reservoir.get(0);// or return reservoir when k > 1;
}
}
``````

#### Random of the index number itself

``````/**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
class Solution {

Note that the head is guaranteed to be not null, so it contains at least one node. */
int count = 0;
ListNode p = root;
while (p != null)
{
p = p.next;
count++;
}
length = count;
}

/** Returns a random node's value. */
public int getRandom() {
Random rand = new Random();
int n = rand.nextInt(length);
ListNode p = root;
while (n-- > 0)
{
p = p.next;
}
return p.val;
}

private int length;
private ListNode root;
}

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