# Leetcode: Linked List Random Node

Similar Problems:

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.

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();
```

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

```## Blog link: https://code.dennyzhang.com/linked-list-random-node
## Basic Ideas: Reservoir Sampling
##      Choose k element from a huge list without knowing the length
##
## Complexity: Time O(n), Space O(1)
##
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

import random
class Solution(object):

"""
Note that the head is guaranteed to be not null, so it contains at least one node.
"""

def getRandom(self):
"""
Returns a random node's value.
:rtype: int
"""
if self.head is None: return None

i = 2
while p:
# Keep with possibility of 1/i
random_v = random.randint(1, i)
# Why we won't quit?
if random_v == 1:
res = p.val
p = p.next
i = i+1
return res

# Your Solution object will be instantiated and called as such: