# 382. Linked List Random Node

#### Medium

***

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.

 

**Example 1:**

![](https://assets.leetcode.com/uploads/2021/03/16/getrand-linked-list.jpg)

```
Input
["Solution", "getRandom", "getRandom", "getRandom", "getRandom", "getRandom"]
[[[1, 2, 3]], [], [], [], [], []]
Output
[null, 1, 3, 2, 2, 3]
Explanation
Solution solution = new Solution([1, 2, 3]);
solution.getRandom(); // return 1
solution.getRandom(); // return 3
solution.getRandom(); // return 2
solution.getRandom(); // return 2
solution.getRandom(); // return 3
// getRandom() should return either 1, 2, or 3 randomly. Each element should have equal probability of returning.
```

 

**Constraints:**

* The number of nodes in the linked list will be in the range `[1, 104]`.
* `-104 <= Node.val <= 104`
* At most `104` calls will be made to `getRandom`.

&#x20;

**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?

```python
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:

    def __init__(self, head: Optional[ListNode]):
        self.head = head
        

    def getRandom(self) -> int:
        result, node, index = self.head, self.head.next, 1
        while node:
            if random.randint(0, index) is 0:
                result = node
            node = node.next
            index += 1
        return result.val


# Your Solution object will be instantiated and called as such:
# obj = Solution(head)
# param_1 = obj.getRandom()
```