# Problem Statement

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.

Implement the `Solution` class:

``````Solution(ListNode head): Initializes the object with the integer array nums.

int getRandom(): Chooses a node randomly from the list and returns its value. All the nodes of the list should be equally likely to be choosen.
``````

Constraints:

• The number of nodes in the linked list will be in the range `[1, 10^4]`.
• `-10^4 <= Node.val <= 10^4`
• At most `10^4` calls will be made to `getRandom`.

## Linear space complexity solution

A naive solution is to store all nodes in an ArrayList and then randomly pick an item:

``````class Solution:
self.nodes = []
while current:
self.nodes.append(current.val)
current = current.next

def getRandom(self) -> int:
return random.choice(self.nodes)
``````

The `getRandom` method has `O(1)` running time. What if the LinkedList is very long? Then `self.nodes` will consume a lot of memory.

## Constant space complexity solution I

We associate each node with a random number and pick the node associated with the maximum number:

``````class Solution:

def getRandom(self) -> int:
current_max = -1
val = None
while current:
ran = random.random()
if ran > current_max:
current_max = ran
val = current.val
current = current.next
return val
``````

## Constant space complexity solution II

Another constant space complexity solution uses Reservoir sampling. It’s a special case with `k = 1`.

``````class Solution:

def getRandom(self) -> int:
res = None
index = 0
while node:
current = random.randint(0, index)
if current == index:
res = node.val
index += 1
node = node.next
return res
``````

One advantange of this solution is it can be easily extended to cases with `k > 1`. For example, one can be asked to generate 5 random nodes. See a clean implementation here.