Linked List Random Node

linked_list_random_node.py

"""
Linked List Random Node
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?

Example:
// Init a singly linked list [1,2,3].
ListNode head = new ListNode(1);
head.next = new ListNode(2);
head.next.next = 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.
solution.getRandom();
"""
from random import random

# 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: ListNode):
        """
        @param head The linked list's head.
        Note that the head is guaranteed to be not null, so it contains at least one node.
        """
        self.numbers = []
        self.get_range(head)
        
    def get_range(self, head: ListNode):
        while head:
            self.numbers.append(head.val)
            head = head.next

    def getRandom(self) -> int:
        """
        Returns a random node's value.
        """
        idx = int(random() * len(self.numbers))
        return self.numbers[idx]


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